# How do you write the trigonometric form into a complex number in standard form 4(cos135+isin135)?

Sep 10, 2016

$4 \left(\cos {135}^{o} + i \sin {135}^{o}\right) = - 2 \sqrt{2} + 2 i \sqrt{2}$

#### Explanation:

As $\cos {135}^{o} = \cos \left({180}^{o} - {45}^{o}\right) = - \cos {45}^{o} = - \frac{1}{\sqrt{2}}$ and

$\sin {135}^{o} = \sin \left({180}^{o} - {45}^{o}\right) = \sin {45}^{o} = \frac{1}{\sqrt{2}}$

Hence $4 \left(\cos {135}^{o} + i \sin {135}^{o}\right)$

= $4 \left(- \frac{1}{\sqrt{2}} + i \times \frac{1}{\sqrt{2}}\right)$

= $- \frac{4}{\sqrt{2}} + 4 \frac{i}{\sqrt{2}}$

But as $\frac{4}{\sqrt{2}} = \frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 4 \frac{\sqrt{2}}{2} = 2 \sqrt{2}$

$4 \left(\cos {135}^{o} + i \sin {135}^{o}\right) = - 2 \sqrt{2} + 2 i \sqrt{2}$