An extension of Imaginary Numbers, complex numbers allow us to combine real numbers with imaginary numbers to produce meaningful results from manipulating and changing things around. This page covers very basic imaginary number ideas. Before starting this, there are some prior knowledge that this page calls upon. Below is a list of the most notable prerequisites that you will need to understand what is going on. Imaginary Numbers The Unit Circle Radians Cubic Functions A complex number is part of the set of complex numbers which we represent as the symbol, C {\displaystyle \displaystyle \mathbb {C} } . Thus any complex number is part of this set C {\displaystyle \displaystyle \mathbb {C} } . Suppose we have complex number, z {\displaystyle \displaystyle z} . Now since its a complex number we can say that, z {\displaystyle \displaystyle z} is part of the set of complex numbers, C {\displaystyle \displaystyle \mathbb {C} } , which we write as z ∈ C {\displaystyle \displaystyle z\in \mathbb {C} } Now the general formula for our complex number, or any complex number, can be written as: z = a + b i {\displaystyle \displaystyle z=a+bi} where a {\displaystyle \displaystyle a} and b {\displaystyle \displaystyle b} are real numbers i.e. a {\displaystyle \displaystyle a} and b {\displaystyle \displaystyle b} in the set of real numbers, R {\displaystyle \displaystyle \mathbb {R} } , which we can again write as a , b ∈ R {\displaystyle \displaystyle a,b\in \mathbb {R} } . The complex number is made up of two parts. They are: The real part, which is a {\displaystyle \displaystyle a} (We can show this as R e a l ( z ) = a {\displaystyle \displaystyle Real(z)=\displaystyle a} ) The imaginary part, the number in front of the i {\displaystyle \displaystyle i} , which is b {\displaystyle \displaystyle b} (We can show this as I m a g i n a r y ( z ) = b {\displaystyle \displaystyle Imaginary(z)=b} ) This is how we connect the real part with the imaginary part. There are different forms of expressing a complex number, but this form is called the Cartesian Form (because its like how we show real numbers on the Cartesian plane ( x , y {\displaystyle x,y} )) An important idea in complex numbers is the conjugate of a complex number. This is simply the opposite of a complex number and is rather easy to figure out. Lets take our previous complex number, z {\displaystyle \displaystyle z} . We already know what this is ( z = a + b i {\displaystyle \displaystyle z=a+bi} ), so the conjugate of it would be: z  ̄ = a − b i {\displaystyle {\bar {z}}=a-bi} We simply change the sign in front of the imaginary part to get our conjugate for z {\displaystyle \displaystyle z} . Pretty simple. The conjugate of z {\displaystyle \displaystyle z} is shown with a special symbol, z  ̄ {\displaystyle \displaystyle {\bar {z}}} . It is the variable for our complex number with a bar over it. This is the conjugate. There are various operations that one can do with complex numbers. Here we demonstrate the 4 basic operations fundamental to the area. Here we suppose that: z 1 = a + b i {\displaystyle z_{1}=a+bi} z 2 = x + y i {\displaystyle z_{2}=x+yi} Adding complex numbers is just like adding two different expressions in algebra. No tricks here. z 1 + z 2 = a + b i + x + y i {\displaystyle \displaystyle z_{1}+z_{2}=a+bi+x+yi} = ( a + x ) + ( b + y ) i {\displaystyle \displaystyle =(a+x)+(b+y)i} Note: We put the real parts in one bracket and the imaginary parts in another. This is just done to make it simpler to understand. In a real question, add them together e.g. z 1 = 1 + i {\displaystyle z_{1}=1+i} and z 2 = 6 + 13 i {\displaystyle z_{2}=6+13i} z 1 + z 2 = 1 + i + 6 + 13 i = ( 1 + 6 ) + ( 1 + 13 ) i {\displaystyle z_{1}+z_{2}=1+i+6+13i=(1+6)+(1+13)i} 7 + 14 i = 7 ( 1 + 2 i ) {\displaystyle 7+14i=7(1+2i)} Subtracting is the opposite process. Be wary of the negative sign however. z 1 − z 2 = ( a + b i ) − ( x + y i ) = a + b i − x − y i {\displaystyle \displaystyle z_{1}-z_{2}=(a+bi)-(x+yi)=a+bi-x-yi} = ( a − x ) + ( b − y ) i {\displaystyle \displaystyle =(a-x)+(b-y)i} e.g. z 1 = 2 − 3 i {\displaystyle z_{1}=2-3i} and z 2 = 4 + i {\displaystyle z_{2}=4+i} z 1 − z 2 = 2 − 3 i − ( 4 + i ) = 2 − 3 i − 4 − i {\displaystyle z_{1}-z_{2}=2-3i-(4+i)=2-3i-4-i} = ( 2 − 4 ) + ( − 3 − 1 ) i = − 2 − 4 i = − 2 ( 1 + 2 i ) {\displaystyle =(2-4)+(-3-1)i=-2-4i=-2(1+2i)} Unlike addition or subtraction, multiplying imaginary numbers isn't as straightforward. We, however, still use the FOIL technique to multiply the two numbers (treating them like expressions). What usually happens is that we multiply an imaginary part by another imaginary part, giving us a negative result. Remember: i 2 = − 1 {\displaystyle \displaystyle i^{2}=-1} z 1 × z 2 = ( a + b i ) × ( x + y i ) = a ( x + y i ) + b i ( x + y i ) {\displaystyle \displaystyle z_{1}\times z_{2}=(a+bi)\times (x+yi)=a(x+yi)+bi(x+yi)} = a x + a y i + b x + b y i 2 = a x + a y i + b x i − b y {\displaystyle \displaystyle =ax+ayi+bx+byi^{2}=ax+ayi+bxi-by} = ( a x − b y ) + ( a y + b x ) i {\displaystyle \displaystyle =(ax-by)+(ay+bx)i} We can show the multiplication of z 1 {\displaystyle z_{1}} and z 2 {\displaystyle z_{2}} as z 1 z 2 . {\displaystyle \displaystyle z_{1}z_{2}.} Remember to be careful when squaring i {\displaystyle \displaystyle i} . Division differs from the previous operations in that it requires knowledge of the conjugate of complex numbers. We can express the division of two complex numbers as: z 1 z 2 = x + y i a + b i {\displaystyle \displaystyle {\frac {z_{1}}{z_{2}}}={\frac {x+yi}{a+bi}}} Using this we must make the denominator (the number at the bottom) a real number. That is we need to multiply the fraction by the conjugate of z 2 {\displaystyle \displaystyle z_{2}} . Here is a step by step process: Find the conjugate of the denominator: In our case, the denominator is z 2 {\displaystyle \displaystyle z_{2}} , which is a + b i {\displaystyle \displaystyle a+bi} . z  ̄ 2 {\displaystyle \displaystyle {\bar {z}}_{2}} is a − b i {\displaystyle \displaystyle a-bi} . That is our conjugate. Multiply the top and bottom by the conjugate: x + y i a + b i × a − b i a − b i {\displaystyle \displaystyle {\frac {x+yi}{a+bi}}\times {\frac {a-bi}{a-bi}}} Multiply like normal: x + y i a + b i × a − b i a − b i = x ( a − b i ) + y i ( a − b i ) a ( a − b i ) + b i ( a − b i ) {\displaystyle \displaystyle {\frac {x+yi}{a+bi}}\times {\frac {a-bi}{a-bi}}={\frac {x(a-bi)+yi(a-bi)}{a(a-bi)+bi(a-bi)}}} Answer: a x − b x i + a y i − b y i 2 a 2 − a b i + a b i − b 2 i 2 = ( a x + b y ) + ( a y − b x ) i a 2 + b 2 {\displaystyle \displaystyle {\frac {ax-bxi+ayi-byi^{2}}{a^{2}-abi+abi-b^{2}i^{2}}}={\frac {(ax+by)+(ay-bx)i}{a^{2}+b^{2}}}} Note the denominator in our answer. It has some significance later. Raising a complex number to a power is done in the same way that we raise a power of an expression in normal algebra, but in keep in mind to be wary of the powers of i {\displaystyle i} . z 1 2 = ( x + y i ) 2 = x 2 + y 2 i 2 + 2 x y i {\displaystyle z_{1}^{2}=(x+yi)^{2}=x^{2}+y^{2}i^{2}+2xyi} = x 2 − y 2 + 2 x y i {\displaystyle =x^{2}-y^{2}+2xyi} Here a few questions to help you remember. Practice makes perfect. Let: z 1 = 3 + 4 i {\displaystyle z_{1}=3+4i} z 2 = 7 + 2 i {\displaystyle z_{2}=7+2i} 1 z 1 + z 2 {\displaystyle z_{1}+z_{2}} . 2 z 1 z 2 {\displaystyle z_{1}z_{2}} 3 z 1 − z 2  ̄ {\displaystyle z_{1}-{\bar {z_{2}}}} 4 ( z 1 − z 2  ̄ ) ( z 2 − z 1  ̄ ) {\displaystyle (z_{1}-{\bar {z_{2}}})(z_{2}-{\bar {z_{1}}})} 5 z 1 z 2  ̄ {\displaystyle {\frac {z_{1}}{\bar {z_{2}}}}} Solving polynomials in the real plane is rather simple and is something covered in Algebra. For example, we can easily solve x 3 − 8 = 0 {\displaystyle x^{3}-8=0} , which is x = 2 {\displaystyle x=2} . However, sometimes we encounter polynomials where there seems to be no real solutions. For example take the quadraticː x 2 + 4 x + 5 = 0 {\displaystyle x^{2}+4x+5=0} Now if we were to use the quadratic formula, a curious error occurs in the discriminant part. Remember the discriminant of an equation is given byː Δ = b 2 − 4 a c {\displaystyle \Delta =b^{2}-4ac} Now if you recall, if Δ < 0 {\displaystyle \Delta <0} , there were no real solutions (The emphasis on 'real' since in the complex plane, there exist solutions). Subbing in our coefficients, we getː Δ = 16 − 4 × 5 × 1 = 16 − 20 = − 4 {\displaystyle \Delta =16-4\times 5\times 1=16-20=-4} Nothing strange here apart from the equation seemingly having no real solutions. When using the quadratic equation we getː x = − 4 ± − 4 2 {\displaystyle x={\frac {-4\pm {\sqrt {-4}}}{2}}} Wait what? − 4 {\displaystyle {\sqrt {-4}}} has no real solutions since you cannot square root a negative. Right? Well in the complex plane you can, since the idea of i {\displaystyle i} is built upon the idea that i = − 1 {\displaystyle i={\sqrt {-1}}} . We can express − 4 {\displaystyle {\sqrt {-4}}} as 4 × − 1 {\displaystyle {\sqrt {4}}\times {\sqrt {-1}}} which is 2 i {\displaystyle 2i} . Now we have solutions for this equationː x = 4 ± 2 i 2 = 2 ± i {\displaystyle x={\frac {4\pm 2i}{2}}=2\pm i} And those our answers. Using the complex plane, we can solve polynomials that have no real solutions. There are a few techniques we can employ to solve polynomials in the complex plane. In the real plane, you cannot really factorizeː a 2 + b 2 {\displaystyle a^{2}+b^{2}} But in the complex plane we can assume that b 2 {\displaystyle b^{2}} can be written as − ( b i ) 2 {\displaystyle -(bi)^{2}} , which when expanded gives usː − b 2 × i 2 = − b 2 × − 1 = b 2 {\displaystyle -b^{2}\times i^{2}=-b^{2}\times -1=b^{2}} . So we can show the sum of two squares asː a 2 − ( b i ) 2 {\displaystyle a^{2}-(bi)^{2}} Which is the difference of two squares soː a 2 + b 2 = a 2 − ( b i ) 2 = ( a + b i ) ( a − b i ) {\displaystyle a^{2}+b^{2}=a^{2}-(bi)^{2}=(a+bi)(a-bi)} e.g. Solve x 2 + 4 {\displaystyle x^{2}+4} = 0 Now we can arrange the equation asː x 2 = − 4 {\displaystyle x^{2}=-4} And then square root it to get our answerː x = ± 2 i {\displaystyle x=\pm 2i} But we can use the sum of perfect squares to also do this x 2 + 4 = ( x 2 + − 4 ) ( x 2 − − 4 ) = ( x + 2 i ) ( x − 2 i ) {\displaystyle x^{2}+4=({\sqrt {x^{2}}}+{\sqrt {-4}})({\sqrt {x^{2}}}-{\sqrt {-4}})=(x+2i)(x-2i)} And soː x = ± 2 i {\displaystyle x=\pm 2i} We can apply this technique of quadratic substitution to simplify polynomials where only even coefficients exist, such as x 4 + x 2 + 4 = 0 {\displaystyle x^{4}+x^{2}+4=0} . Lets use an example to understand how we do thisː z 4 + 2 z 2 − 15 = 0 {\displaystyle z^{4}+2z^{2}-15=0} We can express these coefficients as powers of 2 (hence the quadratic substitution idea). Lets take a variable, say m {\displaystyle m} , and say that m = z 2 {\displaystyle m=z^{2}} ( z 2 ) 2 + 2 ( z 2 ) − 15 = m 2 + 2 m − 15 {\displaystyle (z^{2})^{2}+2(z^{2})-15=m^{2}+2m-15} Now we can solve this like any normal quadraticː m 2 + 2 m − 15 = ( m + 5 ) ( m − 3 ) {\displaystyle m^{2}+2m-15=(m+5)(m-3)} soː m = − 5 , m = 3 {\displaystyle m=-5,m=3} But remember that m = z 2 {\displaystyle m=z^{2}} , so we have one more step to doː z 2 = − 5 , z 2 = 3 {\displaystyle z^{2}=-5,z^{2}=3} And now we square root, remembering to express any negative square root as a positive square root with i {\displaystyle i} ( − 9 = 9 × − 1 = 3 i {\displaystyle {\sqrt {-9}}={\sqrt {9}}\times {\sqrt {-1}}=3i} ) z = ± 5 i , z = ± 3 {\displaystyle z=\pm {\sqrt {5}}i,z=\pm {\sqrt {3}}} Generally used for polynomials with an even number of terms, we can group like terms together (similar to what we do when factoring non-monic trinomials in quadratics). Take the equationː 2 z 3 − 3 z 2 + 16 z − 24 = 0 {\displaystyle 2z^{3}-3z^{2}+16z-24=0} We can group like terms together based on coefficients. Through careful rearranging and finding the HCF we can find thatː 2 z 3 + 16 z − 3 z 2 − 24 = 2 z ( z 2 + 8 ) − 3 ( z 2 + 8 ) = ( 2 z − 3 ) ( z 2 + 8 ) {\displaystyle 2z^{3}+16z-3z^{2}-24=2z(z^{2}+8)-3(z^{2}+8)=(2z-3)(z^{2}+8)} Now we can simply solve using the Null Factor Lawː 2 z − 3 = 0 {\displaystyle 2z-3=0} z = 3 2 {\displaystyle z={\frac {3}{2}}} z 2 = − 8 {\displaystyle z^{2}=-8} z = − 8 = ± 2 2 i {\displaystyle z={\sqrt {-8}}=\pm 2{\sqrt {2}}i} And that's the solutions. Just to make sure you've found all the solutions, take into account what type of polynomial it is (cubic, quadratic, quartic, etc.). Here a cubic will have 3 solutions which we have found. Just like the methods we use when solving polynomials, we can apply the factor theorem to solving polynomials over C {\displaystyle \mathbb {C} } . Remember the factor theoremː If ( x − a ) {\displaystyle (x-a)} is a factor of the polynomial, P ( x ) {\displaystyle P(x)} , then P ( a ) = 0 {\displaystyle P(a)=0} . In complex numbers, we can use this idea to not only prove that something is a factor, but then using division to find a quadratic factor of a polynomial. For example takeː z 3 + i z 2 + 2 i {\displaystyle z^{3}+iz^{2}+2i} and ( z − i ) {\displaystyle (z-i)} is a factor. First we sub in i {\displaystyle i} for z {\displaystyle z} . i 3 + i × i 2 + 2 i = − i × − i ∗ i + 2 i = − 2 i + 2 i = 0 {\displaystyle i^{3}+i\times i^{2}+2i=-i\times -i*i+2i=-2i+2i=0} So we know definitely ( z − i ) {\displaystyle (z-i)} is a factor. Now all we do is divide by this factor (it may be a bit tricky doing this with complex numbers but remember to apply the same rules and watch out for i 2 {\displaystyle i^{2}} ) ( z − i ) ( z 2 + 2 i z − 2 ) = 0 {\displaystyle (z-i)(z^{2}+2iz-2)=0} Now we simply solve the quadratic. Using Complete The Square (same process as with real numbers, but be careful of that i 2 {\displaystyle i^{2}} and also remember that ( a + b i ) 2 = a 2 + 2 a b i − b 2 {\displaystyle (a+bi)^{2}=a^{2}+2abi-b^{2}} ) We end up withː ( z + i ) 2 − 3 = 0 {\displaystyle (z+i)^{2}-3=0} ( z + i ) 2 = 3 {\displaystyle (z+i)^{2}=3} z + i = ± 3 {\displaystyle z+i=\pm {\sqrt {3}}} z = ± 3 − i {\displaystyle z=\pm {\sqrt {3}}-i} And now we have solved our equation. z = i , 3 − i , − 3 − i {\displaystyle z=i,{\sqrt {3}}-i,-{\sqrt {3}}-i} A really major part of solving polynomials over C {\displaystyle \mathbb {C} } is the conjugate root theorem. This idea is unique to complex numbers, thanks to the special properties of i {\displaystyle i} . It is important to note that this idea only applies to polynomials which have REAL coefficients, so x 3 + 2 i x 2 + 3 {\displaystyle x^{3}+2ix^{2}+3} can't really have conjugate complex pairs. The theorem statesː If z = a + b i {\displaystyle z=a+bi} is a root of polynomial, P ( x ) {\displaystyle P(x)} , then so is the conjugate, z  ̄ = a − b i {\displaystyle {\bar {z}}=a-bi} If ( z − a − b i ) {\displaystyle (z-a-bi)} is a factor of polynomial of, P ( x ) {\displaystyle P(x)} , then so is the conjugate factor, ( z − a + b i ) {\displaystyle (z-a+bi)} This has major implications in the way we solve polynomials, causing us to see that every polynomial can be viewed as the product of some factors and the conjugate pairs of two or more conjugate numbers. It also allows us to deduce quickly the other roots of a polynomial without a lengthy process. In complex numbers, we have to find a way to represent these geometrically in order to do various manipulations. The solution to this problem is through Argand diagrams. In general, Argand diagrams can be described as Cartesian planes which have been repurposed to fit the idea of complex numbers. Here, the x {\displaystyle x} -axis becomes the 'real axis' and the y {\displaystyle y} -axis becomes the 'imaginary axis'. Using the Cartesian form, we can plot the locations of our complex numbers on the plane. Suppose we have the complex number, z {\displaystyle z} , be a + b i {\displaystyle a+bi} . We would plot it as in the image shown. Notice how the value for R e ( z ) = a {\displaystyle Re(z)=a} and the value for I m ( z ) = b {\displaystyle Im(z)=b} are plotted. They correspond to a value on their respective axis, a {\displaystyle a} being plotted at the same value as the real part and b {\displaystyle b} plotted as the same value as the imaginary part. The Argand Diagram shares many similarities with the way the Cartesian plane works, in the way that the complex numbers are plotted. In fact, the form of a complex number, a + b i {\displaystyle a+bi} , is called Cartesian or Rectangular form, because of this. There are other forms of complex number representations, one of which is polar form. Polar form is a new way at looking how we describe the location of a point on any plane, not just the Argand Diagram. In Cartesian, we usually describe the location of a point based on two values which when drawn perpendicular from their respective axis, intersect at that point. In polar form however, we describe a point based on its distance from the origin and the angle it has from the positive x-direction. We can apply polar form to many coordinate systems but for now, let's stick to complex numbers. We usually right the coordinates of a complex number in Cartesian form as: a + b i {\displaystyle a+bi} In polar form however, we describe the location of a complex number as the distance from origin and angle (called an argument) from the positive x axis: z = r ( c o s ( θ ) + s i n ( θ ) i ) {\displaystyle z=r(cos(\theta )+sin(\theta )i)} or z = r c i s θ {\displaystyle z=rcis\theta } Now what do all these numbers mean? If you covered unit circles, its evident that c o s ( θ ) {\displaystyle cos(\theta )} can be a representation of the x {\displaystyle x} -coordinate (which in this case is the real number) and s i n ( θ ) {\displaystyle sin(\theta )} can be a representation of the y {\displaystyle y} -coordinate, (which in this case is the imaginary part. Thus that's our first part solved. With r {\displaystyle r} , that's the modulus function, which gives us the distance from origin. We can solve what r {\displaystyle r} is and what θ {\displaystyle \theta } is using the equation: r = a 2 + b 2 {\displaystyle r={\sqrt {a^{2}+b^{2}}}} θ = a r c t a n ( b a ) {\displaystyle \theta =arctan({\frac {b}{a}})} Now, the equation for r {\displaystyle r} looks familiar. Its actually Pythagoras' Theorem, which we are using to determine the length of point from the origin, by taking our side lengths as a {\displaystyle a} and b {\displaystyle b} . We actually did this before and was the multiplication of a complex number and its conjugate: z z  ̄ = ( a + b i ) ( a − b i ) = a 2 + b 2 {\displaystyle z{\bar {z}}=(a+bi)(a-bi)=a^{2}+b^{2}} However, this is taken further and square rooted. We call this the modulus of a function, i.e. the distance it has from the origin. We write this as: | z | = a 2 + b 2 {\displaystyle |z|={\sqrt {a^{2}+b^{2}}}} Notice how the modulus of a number uses the symbol, | x | {\displaystyle |x|} . This is called the absolute value or modulus function, and its definition is consistent with what we did here. Now, the θ {\displaystyle \theta } , is what we call the argument of z {\displaystyle z} and we right this using radians, rather than degrees. To figure out the argument of say, complex number z {\displaystyle z} ( A r g ( z ) {\displaystyle Arg(z)} ), we do the following: Figure out in what quadrant the complex number lies in: We can figure this out by simply looking at the signs of the real and imaginary part of the complex number and relating it to the way that we figure out if an ( x , y ) {\displaystyle (x,y)} coordinate is in what quadrant. Assignment: In our complex number, a + b i {\displaystyle a+bi} , we can relate the trig identities of cos ⁡ θ {\displaystyle \cos \theta } and sin ⁡ θ {\displaystyle \sin \theta } to their appropriate parts (real, imaginary). This is where the knowledge of the unit circle comes in handy, since we can imagine that a {\displaystyle a} and b {\displaystyle b} as x {\displaystyle x} and y {\displaystyle y} , and in a unit circle, x = cos ⁡ θ {\displaystyle x=\cos \theta } and y = sin ⁡ θ {\displaystyle y=\sin \theta } . Through this we can equate to find the argument. Use CAST: Now using CAST we can perform the necessary calculations to find the appropriate angle for our argument. Remember step 1, Figure out what quadrant the complex number lies in? This is where we use it. We can use arctan ⁡ ( b a ) {\displaystyle \arctan({\frac {b}{a}})} but rather we can figure out θ {\displaystyle \theta } by using cos {\displaystyle \cos } and sin {\displaystyle \sin } . To figure out θ {\displaystyle \theta } by doing this we simply do: θ = arccos ⁡ ( a r ) {\displaystyle \theta =\arccos({\frac {a}{r}})} θ = arcsin ⁡ ( b r ) {\displaystyle \theta =\arcsin({\frac {b}{r}})} θ = arctan ⁡ ( b a ) {\displaystyle \theta =\arctan({\frac {b}{a}})} Compare and deduce: Now that we have 3 theta values, we can work out the argument of the complex number is. A positive value means the angle will be in an anti-clockwise direction from 0 whereas a negative angle value is from 0 degrees in a clockwise direction (Remember CAST) Answer: Give the answer in radians. To convert from degrees to radians use the formula: r = ( π 180 ) × θ {\displaystyle r=({\frac {\pi }{180}})\times \theta } Argument of z {\displaystyle z} , has now been found. We can now write our complex number, z {\displaystyle z} as: z = r c i s θ {\displaystyle z=rcis\theta } Converting from cartesian to polar is rather complicated but through these steps you can do so. Take the complex number: z = 1 + i {\displaystyle z=1+i} Remember that | z | = a 2 + b 2 {\displaystyle |z|={\sqrt {a^{2}+b^{2}}}} . Subbing the real, 1 {\displaystyle 1} , and the imaginary, 1 {\displaystyle 1} , values for a {\displaystyle a} and b {\displaystyle b} . So the modulus of z {\displaystyle z} is 1 2 + 1 2 = 2 {\displaystyle {\sqrt {1^{2}+1^{2}}}={\sqrt {2}}} So | z | {\displaystyle |z|} is 2 {\displaystyle {\sqrt {2}}} . Now that we have the modulus, all we need now is the argument, the θ {\displaystyle \theta } . First figure out which quadrant it we can find the complex number in. R e ( z ) {\displaystyle Re(z)} and I m ( z ) {\displaystyle Im(z)} are both positive values so our complex number lies in Quadrant 1. Now we use CAST. Since z {\displaystyle z} is in Quadrant one, the cos {\displaystyle \cos } and sin {\displaystyle \sin } values will be positive. arccos ⁡ ( 1 2 ) = 45 ∘ {\displaystyle \arccos({\frac {1}{\sqrt {2}}})=45^{\circ }} arcsin ⁡ ( 1 2 ) = 45 ∘ {\displaystyle \arcsin({\frac {1}{\sqrt {2}}})=45^{\circ }} So our argument is 45 ∘ {\displaystyle 45^{\circ }} . However, when dealing with polar form, we want to express this as radians. So we can use our formula, π 180 × θ {\displaystyle {\frac {\pi }{180}}\times \theta } . π 180 × 45 ∘ = 45 π 180 = π 4 {\displaystyle {\frac {\pi }{180}}\times 45^{\circ }={\frac {45\pi }{180}}={\frac {\pi }{4}}} So that's our argument, π 4 {\displaystyle {\frac {\pi }{4}}} , in radians. So now we can write out our complex number in polar formː z = 2 c i s π 4 {\displaystyle z={\sqrt {2}}cis{\frac {\pi }{4}}} Conversion from polar to cartesian is that we apply somewhat the opposite process to our polar form. Suppose we have the complex number in polar formː z = 2 c i s ( 7 π 4 ) {\displaystyle z=2cis({\frac {7\pi }{4}})} Now if we use our intuition and some knowledge of the unit circle (its heavily recommended to know about the unit circle before reading this), we find that 7 π 4 {\displaystyle {\frac {7\pi }{4}}} is in fact in the 4th quadrant. So in our cartesian form, I m ( z ) {\displaystyle Im(z)} will be negative and R e ( z ) {\displaystyle Re(z)} will be positive. Now we could just go straight ahead and work out the cos {\displaystyle \cos } and sin {\displaystyle \sin } values for it, but if we use our knowledge of reference angles, we can find that the reference angle for this is π 4 {\displaystyle {\frac {\pi }{4}}} . Now we can go ahead and work out the two parts of z {\displaystyle z} . R e ( z ) = 2 × cos ⁡ 45 ∘ = 2 × 2 2 = 2 {\displaystyle Re(z)=2\times \cos 45^{\circ }=2\times {\frac {\sqrt {2}}{2}}={\sqrt {2}}} I m ( z ) = 2 × − sin ⁡ 45 ∘ = − 2 {\displaystyle Im(z)=2\times -\sin 45^{\circ }=-{\sqrt {2}}} Remember CAST. This is why the sin {\displaystyle \sin } was negative when working out I m ( z ) {\displaystyle Im(z)} . So now our complex number in cartesian form isː z = 2 − 2 i = 2 ( 1 − i ) {\displaystyle z={\sqrt {2}}-{\sqrt {2}}i={\sqrt {2}}(1-i)} In polar form, we are only limited to multiplying and dividing polar forms of complex numbers. Addition or subtraction requires us to convert to cartesian form. For our examples we will be usingː z 1 = r 1 c i s ( θ 1 ) {\displaystyle z_{1}=r_{1}cis(\theta _{1})} z 2 = r 2 c i s ( θ 2 ) {\displaystyle z_{2}=r_{2}cis(\theta _{2})} When multiplying, simply multiply the modulus and add the angles z 1 z 2 = r 1 r 2 c i s ( θ 2 + θ 1 ) {\displaystyle z_{1}z_{2}=r_{1}r_{2}cis(\theta _{2}+\theta _{1})} When dividing we use the same idea, dividing the modulus and subtracting the angles z 1 z 2 = r 1 r 2 c i s ( θ 2 − θ 1 ) {\displaystyle {\frac {z_{1}}{z_{2}}}={\frac {r_{1}}{r_{2}}}cis(\theta _{2}-\theta _{1})} When discussing powers in polar form, we can use De Moivre's Formula (named after French Mathematician Abraham de Moivre) . When raising the complex number, z {\displaystyle z} , to a power in polar form we simply doː z n = r n c i s ( n θ ) {\displaystyle z^{n}=r^{n}cis(n\theta )} Its a simplification of multiplying the complex number multiple times (which was what powers are essentially). Using De Moivre's formula, we can find the roots of a polynomial. Now there are two ways to find roots of a polynomial, either through the rectangular method or the polar form method. Rectangular method uses the idea of squaring a complex number ( x + y i ) {\displaystyle (x+yi)} . Lets take an exampleː Solve z 2 = 5 − 12 i {\displaystyle z^{2}=5-12i} Now, z 2 = 5 − 12 i {\displaystyle z^{2}=5-12i} , however z {\displaystyle z} must've been derived from z = x + y i {\displaystyle z=x+yi} . So lets see what z 2 {\displaystyle z^{2}} looks like if z = x + y i {\displaystyle z=x+yi} z 2 = ( x + y i ) 2 {\displaystyle z^{2}=(x+yi)^{2}} = x 2 − y 2 + 2 x y i {\displaystyle =x^{2}-y^{2}+2xyi} Now we end up with this expanded form. Notice that we can also call this a complex number since there are 2 parts, a real part and an imaginary part. So what we can do, is equate the real and imaginary parts and solve like a normal equation for x {\displaystyle x} and y {\displaystyle y} . x 2 − y 2 = 5 {\displaystyle x^{2}-y^{2}=5} 2 x y = 12 {\displaystyle 2xy=12} We can remove the i {\displaystyle i} since its unnecessary. Now we solve like a simultaneous equation. Start with 2 x y = 12 {\displaystyle 2xy=12} and sub it into x 2 − y 2 = 5 {\displaystyle x^{2}-y^{2}=5} 2 x y = − 12 ⟶ x y = − 6 {\displaystyle 2xy=-12\longrightarrow xy=-6} (divide by 2) y = − 6 x {\displaystyle y={\frac {-6}{x}}} (rearranging it to make y the subject) x 2 − y 2 = 5 → {\displaystyle x^{2}-y^{2}=5\rightarrow } x 2 − ( − 6 x ) 2 = 5 {\displaystyle x^{2}-({\frac {-6}{x}})^{2}=5} (subbing in our known y value) x 2 − 36 x 2 = 5 → x 4 − 36 = 5 x 2 {\displaystyle x^{2}-{\frac {36}{x^{2}}}=5\rightarrow x^{4}-36=5x^{2}} (rationalising everything) x 4 − 5 x 2 − 36 = 0 {\displaystyle x^{4}-5x^{2}-36=0} (we've reached a quartic. We can use quadratic substitution). Let m = x 2 {\displaystyle m=x^{2}} m 2 − 5 m − 36 = 0 → ( m − 9 ) ( m + 4 ) = 0 {\displaystyle m^{2}-5m-36=0\rightarrow (m-9)(m+4)=0} m = 9 , m = − 4 → x 2 = 9 , x 2 = − 4 {\displaystyle m=9,m=-4\rightarrow x^{2}=9,x^{2}=-4} (remember that m = x 2 {\displaystyle m=x^{2}} ) x = ± 3 {\displaystyle x=\pm 3} (remember, x {\displaystyle x} , is our real part of the complex number z {\displaystyle z} so we don't include x 2 = − 4 {\displaystyle x^{2}=-4} which would yield, x = ± 2 i {\displaystyle x=\pm 2i} ) y = ± 2 {\displaystyle y=\pm 2} (using the values of positive and negative x {\displaystyle x} , we find the values of y {\displaystyle y} ) So now we've found the answer (after a very long process) that the roots of z 2 {\displaystyle z^{2}} , areː z = 3 − 2 i {\displaystyle z=3-2i} z = − 3 + 2 i {\displaystyle z=-3+2i} Which we can express asː z = ± ( 3 − 2 i ) {\displaystyle z=\pm (3-2i)} Although cartesian form yields a very comfortable looking answer, it takes a long and tedious process to do so. However, using polar form, we can find the roots rather easily using De Moivre's Formula. Lets explain using an exampleː Solve z 3 = 8 c i s ( 2 π 3 ) {\displaystyle z^{3}=8cis({\frac {2\pi }{3}})} Now De Moivre's theorem states that when raising a number by a powerː z n = r n c i s ( n θ ) {\displaystyle z^{n}=r^{n}cis(n\theta )} If we can use this idea and apply it to our equation The set of complex numbers is denoted C {\displaystyle \mathbb {C} } . A complex number z ∈ C {\displaystyle z\in \mathbb {C} } can be written in Cartesian coordinates as z = a + i b {\displaystyle z=a+ib} where a , b ∈ R {\displaystyle a,b\in \mathbb {R} } . a {\displaystyle a} is called the 'real part' of z {\displaystyle z} and b {\displaystyle b} is called the 'imaginary part' of z {\displaystyle z} . These can also be written in a trigonometric polar form, as z = r ( cos ⁡ ( θ ) + i sin ⁡ ( θ ) ) {\displaystyle z=r(\cos(\vartheta )+i\sin(\vartheta ))} where r ∈ R {\displaystyle r\in \mathbb {R} } is the 'magnitude' of z {\displaystyle z} and θ ∈ [ − π , π ) {\displaystyle \vartheta \in [-\pi ,\pi )} is called the 'argument' of z {\displaystyle z} . These two forms are related by the equations r = a 2 + b 2 θ = arctan ⁡ ( b a ) . {\displaystyle {\begin{aligned}r&={\sqrt {a^{2}+b^{2}}}\\\vartheta &=\arctan \left({\frac {b}{a}}\right).\end{aligned}}} The trigonometric polar form can also be written as z = r e i θ {\displaystyle z=re^{i\vartheta }} by using Euler's Identity e i θ = cos ⁡ ( θ ) + i sin ⁡ ( θ ) . {\displaystyle e^{i\vartheta }=\cos(\vartheta )+i\sin(\vartheta ).} Coordination z = x + y i {\displaystyle z=x+yi} in Cartesian form, z = | z | ( cos ⁡ θ + i sin ⁡ θ ) {\displaystyle z=|z|(\cos \vartheta +i\sin \theta )} in trigonometric polar form, z = | z | e i θ {\displaystyle z=|z|e^{i\vartheta }\,} in polar exponential form. A complex number z ∗ ∈ C {\displaystyle z^{*}\in \mathbb {C} } is a complex conjugate of a number z ∈ C {\displaystyle z\in \mathbb {C} } if and only if z ⋅ z ∗ = 1. {\displaystyle z\cdot z^{*}=1.} If a complex number z {\displaystyle z} is written as z = a + i b {\displaystyle z=a+ib} , then the conjugate is z ∗ = a − i b . {\displaystyle z^{*}=a-ib.} Equivalently in polar form if z = r e i θ {\displaystyle z=re^{i\vartheta }} then z ∗ = r e − i θ = r ( cos ⁡ ( θ ) − i sin ⁡ ( θ ) ) . {\displaystyle z^{*}=re^{-i\vartheta }=r(\cos(\vartheta )-i\sin(\vartheta )).} Operation on complex number and its conjugate z ⋅ z ∗ = | z | ∠ e i θ ⋅ | z | ∠ e − i θ = | z | 2 e i ( θ − θ ) = | z | 2 e 0 = | z | 2 {\displaystyle z\cdot z^{*}=|z|\angle e^{i\vartheta }\cdot |z|\angle e^{-i\vartheta }=|z|^{2}e^{i(\vartheta -\vartheta )}=|z|^{2}e^{0}=|z|^{2}} z z ∗ = | z | e i θ | z | e − i θ = e 2 i θ {\displaystyle {\frac {z}{z^{*}}}={\frac {|z|e^{i\vartheta }}{|z|e^{-i\vartheta }}}=e^{2i\vartheta }} Operation on 2 different complex numbers z 1 ⋅ z 2 = | z 1 | e i θ 1 ⋅ | z 1 | e i θ 2 = | z 1 z 2 | e i ( θ 1 + θ 2 ) {\displaystyle z_{1}\cdot z_{2}=|z_{1}|e^{i\vartheta _{1}}\cdot |z_{1}|e^{i\vartheta _{2}}=|z_{1}z_{2}|e^{i(\vartheta _{1}+\vartheta _{2})}} z 1 z 2 = | z 1 | e i θ 1 | z 2 | e i θ 2 = | z 1 z 2 | e i ( θ 1 − θ 2 ) {\displaystyle {\frac {z_{1}}{z_{2}}}={\frac {|z_{1}|e^{i\vartheta _{1}}}{|z_{2}|e^{i\vartheta _{2}}}}=\left|{\frac {z_{1}}{z_{2}}}\right|e^{i(\vartheta _{1}-\vartheta _{2})}} A careful analysis of the power series for the exponential, sine, and cosine functions reveals the marvelous e i θ = cos ⁡ θ + i sin ⁡ θ {\displaystyle e^{i\theta }=\cos \theta +i\sin \theta \,} of which there is the famous case (for θ = π): e i π = − 1 {\displaystyle e^{i\pi }=-1\,} More generally, x + y i = r ( cos ⁡ θ + i sin ⁡ θ ) = r e i θ {\displaystyle x+yi=r(\cos \theta +i\sin \theta )=re^{i\theta }\,} ( cos ⁡ ( x ) + i sin ⁡ ( x ) ) n = cos ⁡ ( n x ) + i sin ⁡ ( n x ) {\displaystyle (\cos(x)+i\sin(x))^{n}=\cos(nx)+i\sin(nx)} for any real x {\displaystyle x} and integer n {\displaystyle n} . This result is known as . The higher mathematical functions (often called "transcendental functions"), like exponential, log, sine, cosine, etc., can be defined in terms of power series (Taylor series). They can be extended to handle complex arguments in a completely natural way, so these functions are defined over the complex plane. They are in fact "complex analytic functions". Many standard functions can be extended to the complex numbers, and may well be analytic (the most notable exception is the logarithm). Since the power series coefficients of the common functions are real, they work naturally with conjugates. For example: sin ⁡ ( z  ̄ ) = sin ⁡ ( z )  ̄ {\displaystyle \sin({\overline {z}})={\overline {\sin(z)}}\,} log ⁡ ( z  ̄ ) = log ⁡ ( z )  ̄ {\displaystyle \log({\overline {z}})={\overline {\log(z)}}\,} Complex number z = x + y i {\displaystyle z=x+yi} . In Rectangular plane z = z ∠ θ {\displaystyle z=z\angle \theta } . In Polar plane z = z ( cos ⁡ θ + i sin ⁡ θ ) {\displaystyle z=z(\cos \theta +i\sin \theta )} . In trigonometry z = z e j θ {\displaystyle z=ze^{j\theta }\,} . In Complex plane Complex conjugate number z ∗ = x − y i {\displaystyle z^{*}=x-yi} . In Rectangular plane z ∗ = z ∠ − θ {\displaystyle z^{*}=z\angle -\theta } . In Polar plane z ∗ = z ∗ ( cos ⁡ θ − i sin ⁡ θ ) {\displaystyle z^{*}=z^{*}(\cos \theta -i\sin \theta )} . In trigonometry angle z ∗ = z ∗ e j θ {\displaystyle z^{*}=z^{*}e^{j\theta }\,} . In Complex plane Videos Imaginary Numbers are Real [Part 1ː Introduction] by Welch Labs Complex Roots of Polynomials by MathsStatUNSW "Complex Numbers".
{'wiki_source': 'wikiversity', 'other_info': "{'content_model': 'wikitext', 'language': 'en', 'title': 'Complex Numbers', 'type': 'text'}"}