# How do you multiply  (7-3i)(5-i)  in trigonometric form?

Mar 8, 2017

$\left(7 - 3 i\right) \left(5 - i\right) = 2 \sqrt{377} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(- \frac{11}{16}\right)$

#### Explanation:

Let us write the two complex numbers in polar coordinates and let them be

${z}_{1} = {r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ and ${z}_{2} = {r}_{2} \left(\cos \beta + i \sin \beta\right)$

Here, if two complex numbers are ${a}_{1} + i {b}_{1}$ and ${a}_{2} + i {b}_{2}$ ${r}_{1} = \sqrt{{a}_{1}^{2} + {b}_{1}^{2}}$, ${r}_{2} = \sqrt{{a}_{2}^{2} + {b}_{2}^{2}}$ and $\alpha = {\tan}^{- 1} \left({b}_{1} / {a}_{1}\right)$, $\beta = {\tan}^{- 1} \left({b}_{2} / {a}_{2}\right)$

$\left\{{r}_{1} \cdot {r}_{2}\right\} \left\{\left(\cos \alpha + i \sin \alpha\right) \cdot \left(\cos \beta + i \sin \beta\right)\right\}$ or

$\left\{{r}_{1} \cdot {r}_{2}\right\} \left\{\left(\cos \alpha \cos \beta + {i}^{2} \sin \alpha \sin \beta\right) + i \left(\cos \alpha \sin \beta + \cos \beta \sin \alpha\right)\right)$ or

$\left\{{r}_{1} \cdot {r}_{2}\right\} \left\{\left(\cos \alpha \cos \beta - \sin \alpha \sin \beta\right) + i \left(\cos \alpha \sin \beta + \cos \beta \sin \alpha\right)\right)$ or

$\left({r}_{1} \cdot {r}_{2}\right) \cdot \left(\cos \left(\alpha + \beta\right) + i \sin \left(\alpha + \beta\right)\right)$ or

${z}_{1} \cdot {z}_{2}$ is given by $\left({r}_{1} \cdot {r}_{2} , \left(\alpha + \beta\right)\right)$

So for multiplication of complex number ${z}_{1}$ and ${z}_{2}$ , take new angle as $\left(\alpha + \beta\right)$ and modulus os ${r}_{1} \cdot {r}_{2}$ of the modulus of two numbers.

Here $7 - 3 i$ can be written as ${r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ where ${r}_{1} = \sqrt{{7}^{2} + {\left(- 3\right)}^{2}} = \sqrt{58}$ and $\alpha = {\tan}^{- 1} \left(\frac{- 3}{7}\right)$

and $5 - i$ can be written as ${r}_{2} \left(\cos \beta + i \sin \beta\right)$ where ${r}_{2} = \sqrt{{5}^{2} + {\left(- 1\right)}^{2}} = \sqrt{26}$ and $\beta = {\tan}^{- 1} \left(- \frac{1}{5}\right)$

and ${z}_{1} \cdot {z}_{2} = \sqrt{58} \cdot \left(\sqrt{26}\right) \left(\cos \theta + i \sin \theta\right)$, where $\theta = \alpha + \beta$

Hence, $\tan \theta = \tan \left(\alpha + \beta\right) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{\frac{- 3}{7} + \left(- \frac{1}{5}\right)}{1 - \left(\frac{- 3}{7} \times \left(- \frac{1}{5}\right)\right)} = \frac{- \frac{22}{35}}{\frac{32}{35}} = - \frac{22}{32} = - \frac{11}{16}$.

Hence, $\left(7 - 3 i\right) \left(5 - i\right) = \sqrt{58 \times 26} \left(\cos \theta + i \sin \theta\right)$

= $2 \sqrt{377} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(- \frac{11}{16}\right)$