# How do you multiply  (1-2i)(2-3i)  in trigonometric form?

Jul 23, 2016

$\left(1 - 2 i\right) \times \left(2 - 3 i\right) = \sqrt{65} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(\frac{7}{4}\right)$

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

and $2 - 3 i$ can be written as ${r}_{2} \left(\cos \beta + i \sin \beta\right)$ where ${r}_{2} = \sqrt{{2}^{2} + {\left(- 3\right)}^{2}} = \sqrt{4 + 9} = \sqrt{13}$ and $\beta = {\tan}^{- 1} \left(- \frac{3}{2}\right)$

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

Hence, as $\tan \theta = \tan \left(\alpha + \beta\right) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{\left(- 2\right) + \left(- \frac{3}{2}\right)}{1 - \left(- 2\right) \times \left(- \frac{3}{2}\right)} = \frac{- \frac{7}{2}}{1 - 3} = \frac{- \frac{7}{2}}{- 2} = \frac{7}{4}$.

and $z = \sqrt{65}$

Hence, $\left(1 - 2 i\right) \times \left(2 - 3 i\right) = \sqrt{65} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(\frac{7}{4}\right)$

Note that both $\alpha$ and $\beta$ are in fourth quadrant based on signs of sine and cosine functions, hence $\theta = \alpha + \beta$ ought to be between $3 \pi$ and $4 \pi$. Hence as $\tan \theta$ is positive, $\theta$ is in third quadrant.