Why we use conjugate of current than the original phasor in the calculation of Complex Power i-e S=VI*
The voltage and current signals have an angle associated to them, better known as \$\theta_v\$ and \$\theta_i\$, respectively. 

In terms of power, you want the phase difference between those two parameters, an angle we can call '\$\theta\$'. That is, you are looking for:

$$ \theta=\theta_v-\theta_i$$

If you were to find \$P=\text{VI}\$, where \$\text{V}\$ and \$\text{I}\$ have the form \$\text{a}+\text{bi}\$, you are implicitly finding $$ \theta=\theta_v+\theta_i$$

instead of the difference. This can be easily seen if you look at this in terms of Euler's identity. 

Let's say that \$\text{V}\$ and \$\text{I}\$ now have the form \$|{\text{V}}|\angle\theta_v\$ and \$|\text{I}|\angle\theta_i\$.

If you now try to find \$P\$ as \$P=\text{VI}\$, you get 

$$ P=|{\text{V}}||{\text{I}}|\angle(\theta_v+\theta_i)$$

Instead of the correct way:

$$ P=|{\text{V}}||{\text{I}}|\angle(\theta_v-\theta_i)$$

What makes you have the phase difference instead of the sum, is the conjugate of \$\text{I}\$ or \$\text{I}^*\$

When you find the conjugate the magnitude stays the same but the angle has opposite sign. So when you multiply the complex voltage and current, you are also subtracting \$\theta_v\$ and \$\theta_i\$.

Hopefully that clears things up.