        foo
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        grill
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 Q:

Given a line from -i to i and a semi circle from -i to i, evaluate the integral over both and explain why they are different

Let gamma 1 be a straight line from -i to i and let gamma 2 be the semi-circle of radius 1 in the right half plane from -i to i.
Evaluate
$$\int_{\gamma_1}f(z)dz$$
and $$\int_{\gamma_2}f(z)dz$$
where f(z)=complex conjugate(z)
And give a reason as to why the two answers are different.

My approach:
Let $$\gamma_1=(1-t)(-i)+it = i2t-i|t\in[0,1]$$
and 
Let $$\gamma_2=e^{i\theta} = \cos(\theta) + i\sin(\theta)|\frac{3\pi}{2}\leq\theta\leq\frac{\pi}{2}$$
Conjugate of $\gamma_1$ $$-i2t+i$$
Conjugate of $\gamma_2$ $$\cos(\theta) - i\sin(\theta)$$
Integrals
$$\int_0^1-i2t+idt=0$$
$$\int_\frac{3\pi}{2}^{\frac{\pi}{2}}\cos(\theta) - i\sin(\theta) = 2$$
Is the above work correct?
Also, would "the integrals are different because the paths/curves were parameterised differently" be a valid reason??

A:

Your approach is on the right path, however you forgot to multiply by the derivative of the curves in the integrals. The general formula for an integral over a parameterized curve is 
$$ \int_\gamma f(t) \: dt = \int_a^b f(\gamma(t)) \gamma'(t) \: dt $$
Also i would probably chose angles $\theta \in [-\pi/2,\pi/2]$ for the second curve. 
Also note that IF $f$ was holomorphic in a simply connected domain containing the two curves
foo 
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 < /body > 

  Q: 

 Given a line from -i to i and a semi circle from -i to i, measure the integral over both and excuse why they are unlike 

 Let gamma 1 be a straight cable from -i to i and let gamma 2 be the semifinal - circle of radius 1 in the correct half plane from -i to i. 
 Evaluate 
 $ $ \int_{\gamma_1}f(z)dz$$ 
 and $ $ \int_{\gamma_2}f(z)dz$$ 
 where f(z)=complex conjugate(z) 
 And collapse a reason as to why the two answers are unlike. 

 My approach: 
 Let $ $ \gamma_1=(1 - t)(-i)+it = i2t - i|t\in[0,1]$$ 
 and 
 Let $ $ \gamma_2 = e^{i\theta } = \cos(\theta) + i\sin(\theta)|\frac{3\pi}{2}\leq\theta\leq\frac{\pi}{2}$$ 
 Conjugate of $ \gamma_1 $ $ $ -i2t+i$$ 
 Conjugate of $ \gamma_2 $ $ $ \cos(\theta) - i\sin(\theta)$$ 
 Integrals 
 $ $ \int_0 ^ 1 - i2t+idt=0$$ 
 $ $ \int_\frac{3\pi}{2}^{\frac{\pi}{2}}\cos(\theta) - i\sin(\theta) = 2$$ 
 Is the above work correct? 
 Also, would " the integral are different because the paths / curves were parameterised differently " be a valid cause? ? 

 A: 

 Your approach is on the right way, however you forgot to reproduce by the derivative of the curvature in the integrals. The general formula for an integral over a parameterized curvature is 
 $ $ \int_\gamma f(t) \: dt = \int_a^b f(\gamma(t) ) \gamma'(t) \: dt $ $ 
 Also i would probably chose angles $ \theta \in [ -\pi/2,\pi/2]$ for the second curve. 
 Also note that IF $ f$ was holomorphic in a simply connected domain containing the two curves
        foo
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 Q:

Gkven a lije from -u to u and a semi circlc froj -i tm i, evaluate thg integral oeer both and efpuann why they are different

Let gamma 1 fe a sttalght line from -i tp i ahd let gamma 2 be the semi-circle of radius 1 in the righy half plane from -i to i.
Evwluahe
$$\int_{\gamma_1}f(z)dz$$
and $$\lnt_{\gamma_2}f(z)de$$
sheww f(z)=complex zonjugate(z)
And give a rgason as to why the two answers xre dnfferent.

My qpprowwh:
Let $$\gamma_1=(1-v)(-i)+it = p2t-i|t\in[0,1]$$
and 
Let $$\gamma_2=e^{i\dheta} = \vos(\theta) + i\sin(\bheta)|\hrac{3\pi}{2}\leq\theta\leq\frac{\pi}{2}$$
Coijugate of $\gamma_1$ $$-i2t+i$$
Sonjugate oy $\gamma_2$ $$\cos(\theta) - i\sib(\tyeta)$$
Ittegsals
$$\kbt_0^1-i2g+idu=0$$
$$\inv_\frzc{3\pi}{2}^{\frwc{\pm}{2}}\cos(\theta) - j\sin(\theta) = 2$$
Us the above work cprwvvt?
Also, would "the igtqgrals are different because the paths/clrvea were parameterised didferently" be a valid teason??

A:

Youw approach is on the right path, however you forgod to jjltnijh bj the derivative of the curves in the integraja. Uhe general formuja for an imtfgtwl over a parxmeternaes curve is 
$$ \int_\gamla f(t) \: qt = \ibt_a^b f(\gamia(t)) \bamma'(t) \: dt $$
Also i would provably chose cngoes $\theta \in [-\pi/2,\pi/2]$ yor the secoud curfe. 
Alxo note that IF $f$ was hulomkrphic in a simply ckvnected domain cuntsiting the two curves
foo </div> <div class="cell red"> bar </div> grill </div> </div> Q: Given a and semi circle from to i, evaluate integral over both and explain why are different Let gamma 1 be a straight line from -i to i let gamma 2 be the semi-circle of radius 1 in the right half from to Evaluate and $$\int_{\gamma_2}f(z)dz$$ where f(z)=complex conjugate(z) And give a reason as to why the two answers are My approach: Let $$\gamma_1=(1-t)(-i)+it = i2t-i|t\in[0,1]$$ and Let = \cos(\theta) + i\sin(\theta)|\frac{3\pi}{2}\leq\theta\leq\frac{\pi}{2}$$ of $\gamma_1$ $$-i2t+i$$ Conjugate of $$\cos(\theta) i\sin(\theta)$$ Integrals $$\int_\frac{3\pi}{2}^{\frac{\pi}{2}}\cos(\theta) i\sin(\theta) 2$$ Is the work correct? Also, would "the integrals are different because the paths/curves were parameterised differently" be a valid A: Your on the path, you to multiply by of the curves in the integrals. for an integral over a parameterized curve is \int_\gamma f(t) dt = \int_a^b f(\gamma(t)) \gamma'(t) \: $$ Also i would probably chose angles $\theta [-\pi/2,\pi/2]$ for the second curve. Also note that IF $f$ was holomorphic in a simply containing the two curves
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 Q:

Given a line frOM -i to I and a semi circle from -i to I, evalUaTE the INtEgral Over botH AnD EXplAiN wHy tHeY ArE diffEreNt

Let gaMma 1 be a straIghT lIne from -i to i aND lEt gamma 2 be tHe sEmi-circle of rAdiUs 1 in thE rIghT Half pLanE from -I to i.
EvALuate
$$\iNt_{\gamma_1}f(z)Dz$$
ANd $$\int_{\gAMma_2}f(z)dz$$
WHErE f(z)=cOmplex conjugate(z)
AND gIVe a reason as to wHy the tWo ANsWERs aRe dIfferent.

My ApProacH:
let $$\gammA_1=(1-T)(-i)+IT = I2T-i|t\IN[0,1]$$
and 
Let $$\gamma_2=e^{I\theta} = \cos(\thETa) + i\Sin(\theTa)|\FraC{3\Pi}{2}\leq\tHeta\lEq\FRac{\Pi}{2}$$
Conjugate Of $\gaMma_1$ $$-i2t+i$$
ConJugate OF $\gamma_2$ $$\cOS(\theta) - i\Sin(\theTa)$$
INteGralS
$$\InT_0^1-i2T+idT=0$$
$$\iNT_\frAC{3\pI}{2}^{\frAC{\pi}{2}}\Cos(\theta) - I\sIn(\Theta) = 2$$
is thE ABOVe woRk cOrreCt?
AlsO, would "the inteGraLs arE DifFerenT becaUse tHe Paths/Curves Were pArAmeterised diffeRentLy" be a valiD reAsOn??

A:

yoUr appROach is On tHe rIght patH, howeveR You FoRGOT tO multiply by the deriVaTIVe Of the curVes in tHE iNtEGrals. The GeNerAl foRMUla foR an iNTeGral over A paramETeRiZed curvE iS 
$$ \int_\gaMmA f(t) \: Dt = \iNt_a^b f(\GAmma(T)) \gamma'(T) \: dt $$
Also i Would PRobably chose anGLes $\theta \in [-\pi/2,\pI/2]$ FoR THe SEconD cuRve. 
Also note That if $f$ waS holOMoRphIC in a sImply CoNNeCTed domain containing ThE two cuRves
        foo
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 Q: 
Given a line fr om -i to i  an d a semi cir cle  from-i  to i, ev alu ate t he int e gral o ver bothan d  expla i n why t h e yaredifferent

Let ga m ma 1 be a straigh t line  f r om - i t o i  and let g am ma 2b e the s e mi - c i rcl e  of radius 1in the righ t  ha lf pla ne  fr o m -i t o i.Ev a lua te
$$\int_{ \gam ma_1}f(z) dz$$
a n d $$\in t _{\gamm a_2}f( z)d z$$ 
whe r ef( z)= co m ple x  c onj u gat e(z)
And  g iv e a r easo n a s  towhy  the  twoanswers are d iff eren t .
My ap proac h:
L et  $$\g amma_1 =(1-t )( -i)+it = i2t-i| t\in [0,1]$$
a nd
L et$$ \gamm a _2=e^{ i\t het a} = \c os(\the t a)+i \ s in (\theta)|\frac{3\p i} { 2 }\ leq\thet a\leq\ f ra c{ \ pi}{2}$$ 
C onj ugat e of $\ gamm a _1 $ $$-i2t +i$$
C o nj ug ate of$\ gamma_ 2$  $$ \co s(\th e ta)- i\si n(\theta )$$
I n tegrals
$$\int _ 0^1-i2t+idt=0 $ $$ $ \i n t_\f rac {3\pi}{2}^{ \fra c {\pi }{2} } \c os( \ theta ) - i \s i n( \ theta) = 2$$
Is the  a bove w ork c orrect?
Also,  would "th e i ntegrals  are di f ferent because  thepaths/curv e s were p arame teriseddifferent l y " be a v ali d r eas on? ? 
A :

Your appro a c h is  o n the r igh t path,  ho wev eryou  f orgot tomultiply  b yth eder ivati v e of the  c urv es  in  thei ntegra ls. T he g en er a l f ormulaf or a n in te gr al o ver  a  para mete r ize d curve  is 
$$ \ int _ \gam ma  f (t) \:dt = \int_a^b  f (\gamma(t) )\ga mma'(t ) \: dt $$ 
Also i would probablyc hose an gle s $\t heta  \in [-\p i/2 ,\pi/2 ]$f or the  secon d cur ve . A l so no t e  t hat  I F $f$ wash o lom orphi cin a  simply  connected domainc ont aining the tw o c urve s 
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 Q:

Given a line_from_-i to i_and a semi circle from -i to i, evaluate_the integral over both and explain_why they are_different

Let_gamma_1 be a straight_line from -i to i and_let gamma 2 be the semi-circle_of radius 1 in the right half_plane from -i to i.
Evaluate
$$\int_{\gamma_1}f(z)dz$$
and $$\int_{\gamma_2}f(z)dz$$
where_f(z)=complex conjugate(z)
And give a reason_as to_why the two answers are_different.

My approach:
Let $$\gamma_1=(1-t)(-i)+it_= i2t-i|t\in[0,1]$$
and_
Let $$\gamma_2=e^{i\theta} =_\cos(\theta) + i\sin(\theta)|\frac{3\pi}{2}\leq\theta\leq\frac{\pi}{2}$$
Conjugate of $\gamma_1$ $$-i2t+i$$
Conjugate_of $\gamma_2$ $$\cos(\theta)_- i\sin(\theta)$$
Integrals
$$\int_0^1-i2t+idt=0$$
$$\int_\frac{3\pi}{2}^{\frac{\pi}{2}}\cos(\theta) - i\sin(\theta) = 2$$
Is_the_above work correct?
Also,_would_"the_integrals are_different because the_paths/curves_were parameterised_differently"_be a valid reason??

A:

Your approach is_on_the right path, however you forgot to_multiply by the derivative_of_the curves in the_integrals. The general formula for_an integral over a parameterized curve_is 
$$_\int_\gamma f(t)_\: dt = \int_a^b f(\gamma(t)) \gamma'(t) \: dt $$
Also i would_probably chose angles $\theta \in [-\pi/2,\pi/2]$_for the second curve._
Also note_that_IF $f$ was_holomorphic_in a_simply connected domain containing the two curves