# How do you multiply  (4-5i)(2+7i)  in trigonometric form?

Apr 3, 2018

$\textcolor{p u r p \le}{{z}_{1} {z}_{2} = \sqrt{2173} \left[\cos \left(0.396\right) + i \sin \left(0.396\right)\right]}$

#### Explanation:

Convert the numbers to polar form.

$\textcolor{red}{{z}_{1} = 4 - 5 i}$

$\Rightarrow {r}_{1} = \sqrt{{4}^{2} + {\left(- 5\right)}^{2}} = \sqrt{16 + 25} = \sqrt{41}$

$\Rightarrow {\theta}_{1} = \arctan \left(\frac{- 5}{4}\right) \approx 5.387$

$\Rightarrow {r}_{1} \left[\cos {\theta}_{1} + i \sin {\theta}_{1}\right] = \textcolor{red}{\sqrt{41} \left[\cos \left(5.387\right) + i \sin \left(5.387\right)\right]}$

$\textcolor{b l u e}{{z}_{2} = 2 + 7 i}$

$\Rightarrow {r}_{2} = \sqrt{{2}^{2} + {7}^{2}} = \sqrt{4 + 49} = \sqrt{53}$

$\Rightarrow {\theta}_{2} = \arctan \left(\frac{7}{2}\right) \approx 1.292$

$\Rightarrow {r}_{2} \left[\cos {\theta}_{2} + i \sin {\theta}_{2}\right] = \textcolor{b l u e}{\sqrt{53} \left[\cos \left(1.292\right) + i \sin \left(1.292\right)\right]}$

Now to multiply them together, the result will be:

${z}_{1} {z}_{2} = {r}_{1} {r}_{2} \left[\cos \left({\theta}_{1} + {\theta}_{2}\right) + i \sin \left({\theta}_{1} + {\theta}_{2}\right)\right]$

$\Rightarrow {r}_{1} {r}_{2} = \sqrt{41} \sqrt{53} = \sqrt{41 \cdot 53} = \sqrt{2173}$

$\Rightarrow {\theta}_{1} + {\theta}_{2} = \arctan \left(\frac{- 5}{4}\right) + \arctan \left(\frac{7}{2}\right) \approx 6.680$

We usually try to express $\theta$ on the interval

$0 < \theta < 2 \pi$

$6.680 - 2 \pi \approx 0.396$

$\textcolor{p u r p \le}{{z}_{1} {z}_{2} = \sqrt{2173} \left[\cos \left(0.396\right) + i \sin \left(0.396\right)\right]}$