# How do you multiply  (9-4i)(2-i)  in trigonometric form?

Apr 25, 2018

color(red)(z=22.0*[cos(-50.5^@)+i*sin(-50.5^@)]

#### Explanation:

$\text{ }$
Multiply the two complex numbers $\left(9 - 4 i\right) \cdot \left(2 - i\right)$ using $F O I L$ Method.

$\left(9 - 4 i\right) \cdot \left(2 - i\right)$

$\Rightarrow 18 - 9 i - 8 i + 4 {i}^{2}$

$\Rightarrow 18 - 17 i + 4 \cdot \left(- 1\right)$

Note: ${i}^{2} = \left(- 1\right)$

$\Rightarrow 18 - 17 i - 4$

$\Rightarrow 14 - 17 i$

Hence, color(red)((9-4i)*(2-i)=14-17i Intermediate Result 1

Convert this intermediate result to Trigonometric Form.

For a complex number in standard form: color(blue)(z=a+bi,

r=sqrt(a^2+b^2

$\theta = {\tan}^{- 1} \left(\frac{b}{a}\right)$

$\cos \left(\theta\right) = \frac{a}{r}$

$\Rightarrow a = r \cdot \cos \left(\theta\right)$

$\sin \left(\theta\right) = \frac{b}{r}$

$\Rightarrow b = r \cdot \sin \left(\theta\right)$

$\therefore z = r \cdot \left[\cos \left(\theta\right) + i \cdot \sin \left(\theta\right)\right]$

Using Intermediate Result 1,

r=sqrt(a^2+b^2

rArr sqrt(14^2+(-17)^2

rArr sqrt(196+289

rArr sqrt(485

$\therefore r \approx 22.02272$

$\theta = {\tan}^{- 1} \left(\frac{b}{a}\right)$

$\Rightarrow {\tan}^{- 1} \left(\frac{- 17}{14}\right)$

$\theta = {\tan}^{- 1} \left(- 1.21429\right)$

$\theta \approx - {50.52763939}^{\circ}$

$\theta \approx - {50.5}^{\circ}$

Hence,

color(red)(z=22.0*[cos(-50.5^@)+i*sin(-50.5^@)]

Hope it helps.