How do you divide $( i+2) / (7i +3)$ in trigonometric form?

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Answer 1 · 2018-01-05T11:53:38

$sqrt(290)/58(cos(40.24)-isin(40.24))$

For a complex number $z=a+bi$ , we can rewrite it in the form $z=r(costheta+isintheta)$ , where $r=sqrt(a^2+b^2)$ and $theta=tan^(-1)(b/a)$

$z_1=2+i$
$r_1=sqrt(2^2+1^2)=sqrt(5)$
$theta_1=tan^(-1)(1/2)$
$z_1=sqrt(5)(cos(tan^(-1)(1/2))+isin(tan^(-1)(1/2))$

$z_2=3+7i$
$r_2=sqrt(3^2+7^2)=sqrt(58)$
$theta_2=tan^(-1)(7/3)$
$z_2=sqrt(5)(cos(tan^(-1)(7/3))+isin(tan^(-1)(7/3))$

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

$z_1/z_2=sqrt(5)/sqrt(58)(cos(tan^(-1)(1/2)-tan^(-1)(7/3))+isin(tan^(-1)(1/2)-tan^(-1)(7/3)))$
$~~sqrt(290)/58(cos(-40.24)+isin(-40.24))$

Since $cos(x)=cos(-x)$ and $sin(-x)=-sin(x)$

$=sqrt(290)/58(cos(40.24)-isin(40.24))$

How do you divide ( i+2) / (7i +3)( i+2) / (7i +3) in trigonometric form?