How do you simplify (1+i)^2/(-3+2i)^2?

Mar 31, 2017

$\frac{2 i}{5 - 12 i}$

Explanation:

${\left(1 + i\right)}^{2} / {\left(- 3 + 2 i\right)}^{2} = \frac{1 + 2 i + {i}^{2}}{9 - 12 i + 4 {i}^{2}}$

${i}^{2} = - 1$, therefore

$\frac{1 + 2 i + {i}^{2}}{9 - 12 i + 4 {i}^{2}} = \frac{1 + 2 i + \left(- 1\right)}{9 - 12 i + 4 \left(- 1\right)}$

$= \frac{1 + 2 i - 1}{9 - 12 i - 4} = \frac{2 i}{5 - 12 i}$

Mar 31, 2017

The answer is $\textcolor{red}{\frac{2 i}{5 - 12 i}}$.

Explanation:

According to problem(ATP),
$\frac{{\left(1 + i\right)}^{2}}{{\left(- 3 + 2 i\right)}^{2}}$
=$\frac{1 - 1 + 2 i}{4 {\left(i\right)}^{2} - 12 i + 9}$[as ${i}^{2} = - 1$]
=$\frac{2 i}{5 - 12 i}$

Mar 31, 2017

${\left(1 + i\right)}^{2} / {\left(- 3 + 2 i\right)}^{2} = - \frac{24}{169} + \frac{10}{169} i$

Explanation:

Given:

${\left(1 + i\right)}^{2} / {\left(- 3 + 2 i\right)}^{2}$

Let us first simplify:

$\frac{1 + i}{- 3 + 2 i}$

and then square it...

$\frac{1 + i}{- 3 + 2 i} = \frac{\left(1 + i\right) \left(- 3 - 2 i\right)}{\left(- 3 + 2 i\right) \left(- 3 - 2 i\right)}$

$\textcolor{w h i t e}{\frac{1 + i}{- 3 + 2 i}} = \frac{- 3 - 2 i - 3 i - 2 {i}^{2}}{9 - 4 {i}^{2}}$

$\textcolor{w h i t e}{\frac{1 + i}{- 3 + 2 i}} = \frac{- 3 - 2 i - 3 i + 2}{9 + 4}$

$\textcolor{w h i t e}{\frac{1 + i}{- 3 + 2 i}} = \frac{- 1 - 5 i}{13}$

So:

${\left(1 + i\right)}^{2} / {\left(- 3 + 2 i\right)}^{2} = {\left(\frac{1 + i}{- 3 + 2 i}\right)}^{2}$

$\textcolor{w h i t e}{{\left(1 + i\right)}^{2} / {\left(- 3 + 2 i\right)}^{2}} = {\left(\frac{- 1 - 5 i}{13}\right)}^{2}$

$\textcolor{w h i t e}{{\left(1 + i\right)}^{2} / {\left(- 3 + 2 i\right)}^{2}} = \frac{- 24 + 10 i}{169}$

$\textcolor{w h i t e}{{\left(1 + i\right)}^{2} / {\left(- 3 + 2 i\right)}^{2}} = - \frac{24}{169} + \frac{10}{169} i$