# How do you multiply  (-1+8i)(-9+7i)  in trigonometric form?

Jul 23, 2016

$\left(- 1 + 8 i\right) \times \left(- 9 + 7 i\right) = 65 \sqrt{2} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(\frac{79}{47}\right)$

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

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

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

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

{r_1r_2}[(cosalphacosbeta-sinalphasinbeta+i(sinalphacosbeta+sinalphacosbeta)] or

$\left({r}_{1} {r}_{2}\right) \left(\cos \left(\alpha + \beta\right) + i \sin \left(\alpha + \beta\right)\right)$ 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 ${r}_{1} \cdot {r}_{2}$ of the modulus of two numbers.

Here $- 1 + 8 i$ can be written as ${r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ where ${r}_{1} = \sqrt{{\left(- 1\right)}^{2} + {8}^{2}} = \sqrt{65}$ and $\alpha = {\tan}^{- 1} \left(- \frac{8}{1}\right) = {\tan}^{- 1} \left(- 8\right)$

and $- 9 + 7 i$ can be written as ${r}_{2} \left(\cos \beta + i \sin \beta\right)$ where ${r}_{2} = \sqrt{{\left(- 9\right)}^{2} + {7}^{2}} = \sqrt{81 + 49} = \sqrt{130}$ and $\beta = {\tan}^{- 1} \left(\frac{7}{- 9}\right) = {\tan}^{- 1} \left(- \frac{7}{9}\right)$

and ${z}_{1} \cdot {z}_{2} = \sqrt{65} \times \sqrt{130} \left(\cos \theta + i \sin \theta\right)$, where $\theta = \alpha + \beta$

Hence, as $\tan \theta = \tan \left(\alpha + \beta\right) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{\left(- 8\right) + \left(- \frac{7}{9}\right)}{1 - \left(- 8\right) \times \left(- \frac{7}{9}\right)} = \frac{- \frac{79}{9}}{1 - \left(\frac{56}{9}\right)} = \frac{- \frac{79}{9}}{- \frac{47}{9}} = \frac{79}{47}$.

and $z = 65 \sqrt{2}$

Hence, $\left(- 1 + 8 i\right) \times \left(- 9 + 7 i\right) = 65 \sqrt{2} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(\frac{79}{47}\right)$