# How do you multiply  (5+7i)(9-i)  in trigonometric form?

Dec 3, 2017

(5-7i)(9-i)~=77.9(cos(48.1°)+isin(48.1°))

#### Explanation:

First we want to convert the factors into trigonometric form.

We can do this using the formula:
$a + b i = r \left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right)$
where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$ and adjusting $\theta$ to the appropriate quadrant.

$5 + 7 i = {r}_{1} \left(\cos \left({\theta}_{1}\right) + i \sin \left({\theta}_{1}\right)\right)$
r_1=sqrt74~=8.6, theta_1=tan^-1(7/5)~=54.5°

$9 - i = {r}_{2} \left(\cos \left({\theta}_{2}\right) + i \sin \left({\theta}_{2}\right)\right)$
r_2=sqrt(82)~=9.1, theta_2=tan^-1(-1/9)~=-6.3°

If we have two complex numbers in polar form like this,
${z}_{1} = {r}_{1} \left(\cos \left({\phi}_{1}\right) + i \sin \left({\phi}_{1}\right)\right)$
${z}_{2} = {r}_{2} \left(\cos \left({\phi}_{2}\right) + i \sin \left({\phi}_{2}\right)\right)$

the product of them will be:
${z}_{1} \cdot {z}_{2} = {r}_{1} {r}_{2} \left(\cos \left({\phi}_{1} + {\phi}_{2}\right) + i \sin \left({\phi}_{1} + {\phi}_{2}\right)\right)$

Applying this to our numbers, we get:
$\left(5 + 7 i\right) \left(9 - i\right) = {r}_{1} {r}_{2} \left(\cos \left({\theta}_{1} + {\theta}_{2}\right) + i \sin \left({\theta}_{1} + {\theta}_{2}\right)\right)$

r_1r_2=sqrt74sqrt82~=77.9, theta_1+theta_2~=54.5°-6.3°~=48.1°

So, (5-7i)(9-i)~=77.9(cos(48.1°)+isin(48.1°))