# How do you multiply  (4+3i)(-3+2i)  in trigonometric form?

Dec 8, 2016

$\left(4 + 3 i\right) \cdot \left(- 3 + 2 i\right) = 5 \sqrt{13} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(\frac{1}{18}\right)$

#### 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 $\alpha = {\tan}^{- 1} \left({b}_{1} / {a}_{1}\right)$, $\beta = {\tan}^{- 1} \left({b}_{2} / {a}_{2}\right)$

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

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

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

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

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

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

Here $4 + 3 i$ can be written as ${r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ where ${r}_{1} = \sqrt{{4}^{2} + {3}^{2}} = \sqrt{25} = 5$ and $\alpha = {\tan}^{- 1} \frac{3}{4}$

and $- 3 + 2 i$ can be written as ${r}_{2} \left(\cos \beta + i \sin \beta\right)$ where ${r}_{2} = \sqrt{{\left(- 3\right)}^{2} + {2}^{2}} = \sqrt{13}$ and $\beta = {\tan}^{- 1} \left(- \frac{2}{3}\right)$

and ${z}_{1} \cdot {z}_{2} = 5 \sqrt{13} \left(\cos \theta + i \sin \theta\right)$, where $\theta = \alpha + \beta$

Hence, $\tan \theta = \tan \left(\alpha + \beta\right) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{\frac{3}{4} + \left(- \frac{2}{3}\right)}{1 - \frac{3}{4} \times \left(- \frac{2}{3}\right)} = \frac{\frac{1}{12}}{\frac{3}{2}} = \frac{1}{18}$.

Hence, $\left(4 + 3 i\right) \cdot \left(- 3 + 2 i\right) = 5 \sqrt{13} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(\frac{1}{18}\right)$