# How do you write the complex number in trigonometric form -4+4i?

Nov 26, 2016

The answer is $= 4 \sqrt{2} \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)$

#### Explanation:

Let $z = a + i b$ be a complex number.

To convert to trigonometric form

$z = r \left(\cos \theta + i \sin \theta\right)$

We calculate the modulus, ∥z∥=sqrt(a^2+b^2)

$z = - 4 + 4 i$

∥z∥=sqrt(16+16)=sqrt32=4sqrt2

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

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

So, $r = 4 \sqrt{2}$

$\cos \theta = - \frac{1}{\sqrt{2}}$

$\sin \theta = \frac{1}{\sqrt{2}}$

We are in the second quadrant

$\theta = \frac{3 \pi}{4}$

$z = 4 \sqrt{2} \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)$