# What is complex conjugate of  [ (3 + 8i)^4 ] / [ (1+i)^10 ]?

Jan 10, 2016

$- 165 + \frac{721}{32} i$

#### Explanation:

First note that ${\left(1 + i\right)}^{2} = {1}^{2} + 2 i + {i}^{2} = 2 i$ and ${\left(2 i\right)}^{2} = - 4$

So

${\left(1 + i\right)}^{10} = {\left(2 i\right)}^{5} = {\left(2 i\right)}^{2} {\left(2 i\right)}^{2} \left(2 i\right) = \left(- 4\right) \left(- 4\right) \left(2 i\right) = 32 i$

${\left(3 + 8 i\right)}^{4} = {\left(9 + 48 i + 64 {i}^{2}\right)}^{2} = {\left(- 55 + 48 i\right)}^{2}$

$= \left(3025 - 5280 i + 2304 {i}^{2}\right) = 721 - 5280 i$

So:

${\left(3 + 8 i\right)}^{4} / {\left(1 + i\right)}^{10} = \frac{721 - 5280 i}{32 i} = - \frac{721}{32} i - \frac{5280}{32} = - \frac{721}{32} i - 165 = - 165 - \frac{721}{32} i$

So, reversing the sign of the coefficient of $i$, the Complex conjugate is:

$- 165 + \frac{721}{32} i$