How do you divide $\frac{1 - 2 i}{1 - 3 i}$ $(1-2i)/(1-3i)$ in trigonometric form?

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How do you divide $\frac{1 - 2 i}{1 - 3 i}$ $(1-2i)/(1-3i)$ in trigonometric form?

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Answer 1 · 2016-03-25T14:28:30

$\frac{7}{10} + \frac{1}{10} \cdot i$ $7/10+1/10*i$

Explanation:

$\frac{(1 - 2 i) \cdot (1 + 3 i)}{(1 - 3 i) \cdot (1 + 3 i)}$ $((1-2i)*(1+3i))/((1-3i)*(1+3i))$
$= \frac{1 + 3 i - 2 i - 6 i^{2}}{1^{2} - 3 i^{2}}$ $=(1+3i-2i-6i^2)/(1^2-3i^2)$
$= \frac{1 + 3 i - 2 i + 6}{1 + 9}$ $=(1+3i-2i+6)/(1+9)$
$= \frac{7}{10} + \frac{1}{10} \cdot i$ $=7/10+1/10*i$

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How do you divide $\frac{1 - 2 i}{1 - 3 i}$ $(1-2i)/(1-3i)$ in trigonometric form?

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