# What is the trigonometric form of  (2-i)*(1-2i) ?

May 19, 2016

$\left(2 - i\right) \cdot \left(1 - 2 i\right) = - 5 i$

#### Explanation:

Let us first write $\left(2 - i\right)$ and $\left(1 - 2 i\right)$ in trigonometric form.

$a + i b$ can be written in trigonometric form $r {e}^{i \theta} = r \cos \theta + i r \sin \theta = r \left(\cos \theta + i \sin \theta\right)$,
where $r = \sqrt{{a}^{2} + {b}^{2}}$.

Hence $2 - i = \sqrt{{2}^{2} + {\left(- 1\right)}^{2}} \left[\cos \alpha + i \sin \alpha\right]$ or

$\sqrt{5} {e}^{i \alpha}$, where $\cos \alpha = \frac{2}{\sqrt{5}}$ and $\sin \alpha = - \frac{1}{\sqrt{5}}$

$1 - 2 i = \sqrt{{1}^{2} + {\left(- 2\right)}^{2}} \left[\cos \beta + i \sin \beta\right]$ or

$\sqrt{5} {e}^{i \beta}$, where $\cos \beta = \frac{1}{\sqrt{5}}$ and $\sin \beta = \frac{- 2}{\sqrt{5}}$

Hence $\left(2 - i\right) \cdot \left(1 - 2 i\right) = \left(\sqrt{5} {e}^{i \alpha}\right) \times \left(\sqrt{5} {e}^{i \beta}\right) = 5 {e}^{i \left(\alpha + \beta\right)}$

= $5 \left(\cos \left(\alpha + \beta\right) + i \sin \left(\alpha + \beta\right)\right)$

= $5 \left[\left(\cos \alpha \cos \beta - \sin \alpha \sin \beta\right) + i \left(\sin \alpha \cos \beta + \cos \alpha \sin \beta\right)\right]$

= $5 \left[\left(\frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}} - \frac{- 1}{\sqrt{5}} \times \frac{- 2}{\sqrt{5}}\right) + i \left(\frac{- 1}{\sqrt{5}} \times \frac{1}{\sqrt{5}} + \frac{2}{\sqrt{5}} \times \frac{- 2}{\sqrt{5}}\right)\right]$

= $5 \left[\left(\frac{2}{5} - \frac{2}{5}\right) + i \left(- \frac{1}{5} - \frac{4}{5}\right)\right]$ = $5 \times - \frac{5}{5} i = - 5 i$