# How do you divide  (5+2i)/(7+i)  in trigonometric form?

Apr 10, 2017

Division of two complex numbers in trigonometric form is defined as:

$\frac{{r}_{1} \left(\cos \left({\theta}_{1}\right) + i \sin \left({\theta}_{1}\right)\right)}{{r}_{2} \left(\cos \left({\theta}_{2}\right) + i \sin \left({\theta}_{2}\right)\right)} = {r}_{1} / {r}_{2} \left(\cos \left({\theta}_{1} - {\theta}_{2}\right) + i \sin \left({\theta}_{1} - {\theta}_{2}\right)\right)$

#### Explanation:

Given: $\frac{5 + 2 i}{7 + i}$

We need to convert the dividend and the divisor into trigonometric form.

${r}_{1} = \sqrt{{a}_{1}^{2} + {b}_{1}^{2}}$

${r}_{1} = \sqrt{{5}^{2} + {2}^{2}}$

${r}_{1} = \sqrt{25 + 4}$

${r}_{1} = \sqrt{29}$

Because the signs of "a" and "b" are in the 1st quadrant, we use the equation:

${\theta}_{1} = {\tan}^{-} 1 \left({b}_{1} / {a}_{1}\right)$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{2}{5}\right)$

Note: We would have add $\pi$ for the 2nd or 3rd quadrant and $2 \pi$ for the 4th.

${r}_{2} = \sqrt{{a}_{2}^{2} + {b}_{2}^{2}}$

${r}_{2} = \sqrt{{7}^{2} + {1}^{2}}$

${r}_{2} = \sqrt{49 + 1}$

${r}_{2} = \sqrt{50}$

Because the signs of "a" and "b" are in the 1st quadrant, we use the equation:

${\theta}_{2} = {\tan}^{-} 1 \left({b}_{2} / {a}_{2}\right)$

${\theta}_{2} = {\tan}^{-} 1 \left(\frac{1}{7}\right)$

$\frac{5 + 2 i}{7 + i} =$

$\frac{\sqrt{29} \left(\cos \left({\tan}^{-} 1 \left(\frac{2}{5}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{2}{5}\right)\right)\right)}{\sqrt{50} \left(\cos \left({\tan}^{-} 1 \left(\frac{1}{7}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{1}{7}\right)\right)\right)} =$

$\sqrt{\frac{29}{50}} \left(\cos \left({\tan}^{-} 1 \left(\frac{2}{5}\right) - {\tan}^{-} 1 \left(\frac{1}{7}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{2}{5}\right) - {\tan}^{-} 1 \left(\frac{1}{7}\right)\right)\right)$