How do you divide $\frac{5 - 12 i}{6 + 2 i}$ $(5-12i)/(6+2i)$ in trigonometric form?

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How do you divide $\frac{5 - 12 i}{6 + 2 i}$ $(5-12i)/(6+2i)$ in trigonometric form?

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Answer 1 · 2016-02-02T15:01:09

$\frac{3}{20} - \frac{41 i}{20}$ $3/20-(41i)/20$

Explanation:

Multiply the expression with the cojugate of 6+2i, both in the numerator and the denominator. Thus it is,

$\frac{5 - 12 i}{6 + 2 i} \cdot \frac{6 - 2 i}{6 - 2 i}$ $(5-12i)/ (6+2i) * (6-2i)/(6-2i)$ = $\frac{6 - 82 i}{40}$ $(6-82i)/40$ = $\frac{3}{20} - \frac{41 i}{20}$ $3/20-(41i)/20$

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How do you divide $\frac{5 - 12 i}{6 + 2 i}$ $(5-12i)/(6+2i)$ in trigonometric form?

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Explanation:

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Science

Math

Humanities

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How do you divide 5−12i6+2i (5-12i)/(6+2i) in trigonometric form?

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Explanation:

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How do you divide $\frac{5 - 12 i}{6 + 2 i}$ $(5-12i)/(6+2i)$ in trigonometric form?