How do you divide $(-3-4i)/(5+2i)$ in trigonometric form?

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Answer 1 · 2018-04-26T07:37:03

$5/sqrt(29)(cos(0.540)+isin(0.540))~~0.79+0.48i$

$(-3-4i)/(5+2i)=-(3+4i)/(5+2i)$

$z=a+bi$ can be written as $z=r(costheta+isintheta)$ , where

For $z_1=3+4i$ :
$r=sqrt(3^2+4^2)=5$
$theta=tan^-1(4/3)=~~0,927$

For $z_2=5+2i$ :
$r=sqrt(5^2+2^2)=sqrt29$
$theta=tan^-1(2/5)=~~0.381$

For $z_1/z_2$ :
$z_1/z_2=r_1/r_2(cos(theta_1-theta_2)+isin(theta_1-theta_2))$

$z_1/z_2=5/sqrt(29)(cos(0.921-0.381)+isin(0.921-0.381))$

$z_1/z_2=5/sqrt(29)(cos(0.540)+isin(0.540))=0.79+0.48i$

Proof:
$-(3+4i)/(5+2i)*(5-2i)/(5-2i)=-(15+20i-6i+8)/(25+4)=(23+14i)/29=0.79+0.48i$

How do you divide (-3-4i)/(5+2i) (-3-4i)/(5+2i) in trigonometric form?