# Solve 1/1+i in polar form ?

Mar 28, 2018

$\frac{1}{1 + i} = \frac{\sqrt{2}}{2} \left[\cos \left(- \frac{\pi}{4}\right) + i \sin \left(- \frac{\pi}{4}\right)\right]$

#### Explanation:

$\frac{1}{1 + i}$

and express the result in the form $r \left[\cos \theta + i \sin \theta\right]$.

First convert the numerator and denominator to polar form.

$\frac{1}{1 + i} = \frac{1 \left[\cos \left(0\right) + i \sin \left(0\right)\right]}{\sqrt{2} \left[\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right]}$

Now we can find the result like so:

$\Rightarrow \frac{1}{\sqrt{2}} \left[\cos \left(0 - \frac{\pi}{4}\right) + i \sin \left(0 - \frac{\pi}{4}\right)\right]$

$\Rightarrow \frac{\sqrt{2}}{2} \left[\cos \left(- \frac{\pi}{4}\right) + i \sin \left(- \frac{\pi}{4}\right)\right]$

Mar 28, 2018

1/sqrt(2)(cos(-1/4pi)+isin(-1/4pi)

#### Explanation:

The method Sharkbasket used is completly right, this is just a second way...
$\frac{1}{1 + i} \to$
$\text{Expand the fraction by} \left(1 - i\right)$
$\text{to get rid of the i in the denominator}$

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

From here on it is basicly the same...

$r = \sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{1}{2}\right)}^{2}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$
$\theta = {\tan}^{-} 1 \left(\frac{\frac{1}{2}}{- \frac{1}{2}}\right) = {\tan}^{-} 1 \left(- 1\right) = - \frac{1}{4} \pi$

1/sqrt(2)(cos(-1/4pi)+isin(-1/4pi)

Mar 28, 2018

see a solution process below;

#### Explanation:

$\frac{1}{1} + i$

$1$ in polar form $= 1 \left(\cos 0 + i \sin 0\right)$

Where;

$1 = r = \sqrt{1 + 0} = 1$

theta = tan-¹0 =0

r=√(1+1) =√2

theta = tan-¹(1/1)=45

Now;

$\frac{1}{1} + i = \frac{1 \left(\cos 0 + i \sin 0\right)}{\sqrt{2} \left(\cos 45 + i \sin 45\right)}$

By complex analysis we have;

$\frac{1}{\sqrt{2}} \left[\cos \left(0 - 45\right) + i \sin \left(0 - 45\right)\right]$

Therefore;

$\frac{1}{1} + i = \frac{\sqrt{2}}{2} \left[\cos 45 - i \sin 45\right]$

Note: $\cos \left(- \theta\right) = \cos \theta$