How do you write the complex number in trigonometric form $5 + 2 i$ $5+2i$ ?

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Answer 1 · 2016-11-19T16:54:43

$5 + 2 i = \sqrt{29} cos θ + i \sqrt{29} sin θ$ $5+2i=sqrt29costheta+isqrt29sintheta$ , where $θ = {tan}^{- 1} (\frac{2}{5})$ $theta=tan^(-1)(2/5)$

A number $a + i b$ $a+ib$ can be written in trigonometric form as

$r cos θ + i r sin θ$ $rcostheta+irsintheta$ .

As $r cos θ = a$ $rcostheta=a$ and $r sin θ = b$ $rsintheta=b$ , squaring and adding them we get $r^{2} = a^{2} + b^{2}$ $r^2=a^2+b^2$ .

As such for $5 + 2 i$ $5+2i$ , $r = \sqrt{5^{2} + 2^{2}} = \sqrt{25 + 4} = \sqrt{29}$ $r=sqrt(5^2+2^2)=sqrt(25+4)=sqrt29$

and $cos θ = \frac{5}{\sqrt{29}}$ $costheta=5/sqrt29$ and $sin θ = \frac{2}{\sqrt{29}}$ $sintheta=2/sqrt29$

i.e. $tan θ = \frac{2}{5}$ $tantheta=2/5$ and $θ = {tan}^{- 1} (\frac{2}{5})$ $theta=tan^(-1)(2/5)$

Hence in trigonometric form

$5 + 2 i = \sqrt{29} cos θ + i \sqrt{29} sin θ$ $5+2i=sqrt29costheta+isqrt29sintheta$ , where $θ = {tan}^{- 1} (\frac{2}{5})$ $theta=tan^(-1)(2/5)$

How do you write the complex number in trigonometric form 5+2i5+2i5+2i?