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

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

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Answer 1 · 2016-02-12T06:51:30

$2 \sqrt{1189} (cos (α + β) + i sin (α + β))$ $2sqrt(1189)(cos (alpha+beta)+isin(alpha+beta))$ , where $α = {tan}^{- 1} (- \frac{1}{9})$ $alpha=tan^(-1)(-1/9)$ and $β = {tan}^{- 1} (- \frac{3}{7})$ $beta=tan^(-1)(-3/7)$

Explanation:

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

Using this $(9 - i) = \sqrt{82} (cos α + i sin α)$ $(9-i)=sqrt82(cos alpha+i sin alpha)$ , and $α = {tan}^{- 1} (- \frac{1}{9})$ $alpha=tan^(-1)(-1/9)$

Similarly, $(7 - 3 i) = \sqrt{58} (cos β + i sin β)$ $(7-3i)=sqrt58(cos beta+i sin beta)$ , and $β = {tan}^{- 1} (- \frac{3}{7})$ $beta=tan^(-1)(-3/7)$

Multiplication of $(\sqrt{82} (cos α + i sin α)$ $(sqrt82(cos alpha+i sin alpha)$ and $\sqrt{58} (cos β + i sin β)$ $sqrt58(cos beta+i sin beta)$ is given by

$2 \sqrt{1189} (cos (α + β) + i sin (α + β))$ $2sqrt(1189)(cos (alpha+beta)+isin(alpha+beta))$ , where $α = {tan}^{- 1} (- \frac{1}{9})$ $alpha=tan^(-1)(-1/9)$ and $β = {tan}^{- 1} (- \frac{3}{7})$ $beta=tan^(-1)(-3/7)$ , as $\sqrt{82 \cdot 58} = 2 \sqrt{1189}$ $sqrt(82*58)=2sqrt(1189)$ .

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

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

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Explanation:

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