# How do I calculate the Real and Imaginary Parts of this equation?

## $z = {\left({e}^{2 + i \frac{\pi}{2}} / \left(1 + 3 i\right)\right)}^{2}$

Feb 11, 2018

$\text{Real part } = 0.08 \cdot {e}^{4}$
$\text{and Imaginary part } = 0.06 \cdot {e}^{4}$

#### Explanation:

$\exp \left(a + b\right) = {e}^{a + b} = {e}^{a} \cdot {e}^{b} = \exp \left(a\right) \cdot \exp \left(b\right)$
$\exp \left(i \theta\right) = \cos \left(\theta\right) + i \sin \left(\theta\right)$
$\implies {e}^{2 + i \cdot \frac{\pi}{2}} = {e}^{2} \cdot \exp \left(i \cdot \frac{\pi}{2}\right) = {e}^{2} \cdot \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right)$
$= {e}^{2} \cdot \left(0 + i\right) = {e}^{2} \cdot i$

$\frac{1}{1 + 3 i} = \frac{1 - 3 i}{\left(1 - 3 i\right) \left(1 + 3 i\right)} = \frac{1 - 3 i}{10} = 0.1 - 0.3 i$

$\text{So we have}$

${\left({e}^{2} \cdot i \cdot \left(0.1 - 0.3 i\right)\right)}^{2}$
$= {e}^{4} \cdot \left(- 1\right) \cdot {\left(0.1 - 0.3 \cdot i\right)}^{2}$
$= - {e}^{4} \cdot \left(0.01 + 0.09 \cdot {i}^{2} - 2 \cdot 0.1 \cdot 0.3 \cdot i\right)$
$= - {e}^{4} \cdot \left(- 0.08 - 0.06 \cdot i\right)$
$= {e}^{4} \left(0.08 + 0.06 \cdot i\right)$

$\implies \text{Real part } = 0.08 \cdot {e}^{4}$
$\text{and Imaginary part } = 0.06 \cdot {e}^{4}$

Feb 11, 2018

$R l \left(z\right) = \frac{2}{25} {e}^{4} , \mathmr{and} , I m \left(z\right) = \frac{3}{50} {e}^{4}$.

#### Explanation:

Recall that, ${e}^{i \theta} = \cos \theta + i \sin \theta \ldots \ldots \ldots \ldots . . \left(\square\right)$.

$\therefore z = {\left(\frac{{e}^{2 + i \frac{\pi}{2}}}{1 + 3 i}\right)}^{2}$,

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

$= {e}^{2 \cdot \left(2 + i \frac{\pi}{2}\right)} / {\left(1 + 3 i\right)}^{2}$,

$= {e}^{4 + i \pi} / {\left(1 + 3 i\right)}^{2}$,

$= \frac{{e}^{4} \cdot {e}^{i \pi}}{1 + 3 i} ^ 2$,

$= \frac{{e}^{4} \cdot \left(\cos \pi + i \sin \pi\right)}{1 + 3 i} ^ 2$,

$= \frac{{e}^{4} \left(- 1 + i \cdot 0\right)}{1 + 3 i} ^ 2$,

$= - {e}^{4} \cdot \frac{1}{1 + 3 i} ^ 2 \cdot {\left(1 - 3 i\right)}^{2} / {\left(1 - 3 i\right)}^{2}$,

$= - \frac{{e}^{4} {\left(1 - 3 i\right)}^{2}}{\left(1 + 3 i\right) \left(1 - 3 i\right)} ^ 2$,

$= - \frac{{e}^{4} {\left(1 - 3 i\right)}^{2}}{1 - 9 {i}^{2}} ^ 2$,

$= - \frac{{e}^{4} \left(1 - 6 i + 9 {i}^{2}\right)}{1 - 9 \left(- 1\right)} ^ 2$,

$= - \frac{{e}^{4} \left(1 - 6 i - 9\right)}{10} ^ 2$,

$= - \frac{{e}^{4} \left(- 8 - 6 i\right)}{100}$,

$= \frac{{e}^{4} \left(4 + 3 i\right)}{50}$.

$\Rightarrow R l \left(z\right) = \frac{2}{25} {e}^{4} , \mathmr{and} I m \left(z\right) = \frac{3}{50} {e}^{4}$.

Feb 11, 2018

$\setminus$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus {\left(\frac{{e}^{2 + i \setminus \frac{\pi}{2}}}{1 + 3 i}\right)}^{2} \setminus = \setminus \frac{2 {e}^{4}}{25} + \frac{3 {e}^{4}}{50} i .$

#### Explanation:

$\setminus$

$\text{We will work this out, working on the complex exponential}$
$\text{part first.}$

$\text{Here we go: }$

${\left(\frac{{e}^{2 + i \setminus \frac{\pi}{2}}}{1 + 3 i}\right)}^{2} \setminus = \setminus {\left({e}^{2 + i \setminus \frac{\pi}{2}}\right)}^{2} / {\left(1 + 3 i\right)}^{2} \setminus = \setminus \frac{{e}^{4 + i \setminus \pi}}{1 + 3 i} ^ 2 \setminus = \setminus \frac{{e}^{4} {e}^{i \setminus \pi}}{1 + 3 i} ^ 2$

$\setminus q \quad \setminus q \quad \setminus q \quad = \setminus \frac{{e}^{4} \left(\cos \left(\setminus \pi\right) + i \sin \left(\setminus \pi\right)\right)}{1 + 3 i} ^ 2 \setminus = \setminus \frac{{e}^{4} \left(- 1 + i \setminus \cdot 0\right)}{1 + 3 i} ^ 2$

$\setminus q \quad \setminus q \quad \setminus q \quad = \setminus {e}^{4} \setminus \cdot \frac{- 1}{1 + 3 i} ^ 2 \setminus = \setminus {e}^{4} \setminus \cdot \frac{- 1}{1 + 3 i} ^ 2 \setminus \cdot {\left(1 - 3 i\right)}^{2} / {\left(1 - 3 i\right)}^{2}$

$\setminus q \quad \setminus q \quad \setminus q \quad = \setminus {e}^{4} \setminus \cdot \frac{- 1 \setminus \cdot {\left(1 - 3 i\right)}^{2}}{{\left(1 + 3 i\right)}^{2} {\left(1 - 3 i\right)}^{2}} \setminus = \setminus {e}^{4} \setminus \cdot \frac{- 1 \setminus \cdot {\left(1 - 3 i\right)}^{2}}{{\left[\left(1 + 3 i\right) \left(1 - 3 i\right)\right]}^{2}}$

$\setminus q \quad \setminus q \quad \setminus q \quad = \setminus {e}^{4} \setminus \cdot \frac{- 1 \setminus \cdot \left(1 - 6 i + 9 {i}^{2}\right)}{{1}^{2} + {3}^{2}} ^ 2 \setminus = \setminus {e}^{4} \setminus \cdot \frac{- 1 \setminus \cdot \left(1 - 6 i - 9\right)}{10} ^ 2$

$\setminus q \quad \setminus q \quad \setminus q \quad = \setminus {e}^{4} \setminus \cdot \frac{- 1 \setminus \cdot \left(- 8 - 6 i\right)}{100} = \setminus {e}^{4} \setminus \cdot \frac{8 + 6 i}{100}$

$\setminus q \quad \setminus q \quad \setminus q \quad = \setminus {e}^{4} \cdot \textcolor{red}{\cancel{2}} \cdot \frac{4 + 3 i}{\textcolor{red}{\cancel{2}} \setminus \cdot 50} \setminus = \setminus {e}^{4} \cdot \left(\frac{4}{50} + \frac{3}{50} i\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad = \setminus {e}^{4} \cdot \left(\frac{2}{25} + \frac{3}{50} i\right) \setminus = \setminus \frac{2 {e}^{4}}{25} + \frac{3 {e}^{4}}{50} i .$

$\setminus$

$\text{Thus:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus {\left(\frac{{e}^{2 + i \setminus \frac{\pi}{2}}}{1 + 3 i}\right)}^{2} \setminus = \setminus \frac{2 {e}^{4}}{25} + \frac{3 {e}^{4}}{50} i .$