How do you multiply $(9-i)(7-3i)$ in trigonometric form?

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Answer 1 · 2016-02-12T06:51:30

$2sqrt(1189)(cos (alpha+beta)+isin(alpha+beta))$ , where $alpha=tan^(-1)(-1/9)$ and $beta=tan^(-1)(-3/7)$

A complex number of form $a+bi$ can be written as $r(cos theta +i sin theta)$ , where $r=sqrt(a^2+b^2)$ and $theta=tan^(-1)(b/a)$ .

Using this $(9-i)=sqrt82(cos alpha+i sin alpha)$ , and $alpha=tan^(-1)(-1/9)$

Similarly, $(7-3i)=sqrt58(cos beta+i sin beta)$ , and $beta=tan^(-1)(-3/7)$

Multiplication of $(sqrt82(cos alpha+i sin alpha)$ and $sqrt58(cos beta+i sin beta)$ is given by

$2sqrt(1189)(cos (alpha+beta)+isin(alpha+beta))$ , where $alpha=tan^(-1)(-1/9)$ and $beta=tan^(-1)(-3/7)$ , as $sqrt(82*58)=2sqrt(1189)$ .

How do you multiply (9-i)(7-3i) (9-i)(7-3i) in trigonometric form?