What is the trigonometric form of $(5 - i) \cdot (3 + i)$ $(5-i)*(3+i)$ ?

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Answer 1 · 2017-04-25T14:20:55

$2 \sqrt{65} \cdot (cos θ + i sin θ)$ $2sqrt65 * ( cos theta + isin theta)$

where, $tan θ = \frac{1}{8} or θ = {tan}^{- 1} (\frac{1}{8})$ $tan theta = 1/8 or theta = tan^-1(1/8)$

$(5 - i) \cdot (3 + i)$ $(5-i)*(3+i)$

$= 5 \cdot 3 + 5 i - 3 i - i \cdot i$ $= 5*3+5i-3i-i*i$

$= 15 + 2 i - \sqrt{- 1} \cdot \sqrt{- 1} = 15 + 2 i - (- 1)$ $=15+2i-sqrt(-1)*sqrt(-1)=15+2i-(-1)$

$= 15 + 2 i + 1$ $=15+2i+1$

$= 16 + 2 i$ $=16+2i$

$= \frac{16 + 2 i}{\sqrt{16^{2} + 2^{2}}} \cdot \sqrt{16^{2} + 2^{2}}$ $=(16+2i)/(sqrt(16^2+2^2))*sqrt(16^2+2^2)$

$= \frac{16 + 2 i}{\sqrt{260}} \cdot \sqrt{260}$ $=[16+2i]/[sqrt(260)]*sqrt(260)$

$= 2 \sqrt{65} [\frac{16 + 2 i}{2 \sqrt{65}}]$ $=2sqrt65[(16+2i)/(2sqrt65)]$

$= 2 \sqrt{65} [\frac{8}{\sqrt{65}} + \frac{i}{\sqrt{65}}]$ $=2sqrt65[8/sqrt65+i/sqrt65]$

$= 2 \sqrt{65} [\frac{8}{\sqrt{65}} + i \cdot \frac{1}{\sqrt{65}}]$ $=2sqrt65[8/sqrt65+i*1/sqrt65]$

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$= 2 \sqrt{65} \cdot (cos θ + i sin θ)$ $=2sqrt65 * ( cos theta + isin theta)$

where, $tan θ = \frac{1}{8} or θ = {tan}^{- 1} (\frac{1}{8})$ $tan theta = 1/8 or theta = tan^-1(1/8)$

What is the trigonometric form of (5-i)*(3+i) (5−i)⋅(3+i) (5-i)*(3+i) ?