# How do you divide ( 4i+1) / (6i +5 ) in trigonometric form?

Jun 12, 2016

To convert non-zero complex number $a + i b$ into trigonometric form $r \left(\cos \alpha + i \sin \alpha\right)$
we have to multiply and divide it by $\sqrt{{a}^{2} + {b}^{2}}$ getting
$\sqrt{{a}^{2} + {b}^{2}} \left(\frac{a}{\sqrt{{a}^{2} + {b}^{2}}} + i \frac{b}{\sqrt{{a}^{2} + {b}^{2}}}\right)$
Now there is always one and only one angle $\alpha$ such that
$\cos \alpha = \frac{a}{\sqrt{{a}^{2} + {b}^{2}}}$ and
$\sin \alpha = \frac{b}{\sqrt{{a}^{2} + {b}^{2}}}$

So, in trigonometric form our number would look like
$\sqrt{{a}^{2} + {b}^{2}} \left(\cos \alpha + i \sin \alpha\right)$
where angle $\alpha$ is defined by its $\cos$ and $\sin$ as explained above.

Furthermore, trigonometric form $\cos \alpha + i \sin \alpha$ is, using the Euler's formula, equivalent to ${e}^{i \alpha}$, which will make it easy to multiply and divide complex numbers.

In our problem we have, using this logic,
$4 i + 1 = \sqrt{17} \left(\cos \phi + i \sin \phi\right) = \sqrt{17} {e}^{i \phi}$
where $\phi = \arccos \left(\frac{1}{\sqrt{17}}\right) \approx {75.96}^{o}$

$6 i + 5 = \sqrt{61} \left(\cos \psi + i \sin \psi\right) = \sqrt{61} {e}^{i \psi}$
where $\psi = \arccos \left(\frac{5}{\sqrt{61}}\right) \approx {50.19}^{o}$

Therefore,
$\frac{4 i + 1}{6 i + 5} = \frac{\sqrt{17}}{\sqrt{61}} {e}^{i \phi} / {e}^{i \psi}$

$= \sqrt{\frac{17}{61}} {e}^{i \left(\phi - \psi\right)}$
$= \sqrt{\frac{17}{61}} \left(\cos \left(\phi - \psi\right) + i \sin \left(\phi - \psi\right)\right)$
$\approx \sqrt{\frac{17}{61}} \left(\cos \left({25.77}^{o}\right) + i \sin \left({25.77}^{o}\right)\right)$
$\approx 0.5279 \left(0.9005 + i \cdot 0.4347\right)$
$\approx 0.4754 + i \cdot 0.2295$