How do you add $(7+4i)$ and $(2+6i)$ in trigonometric form?

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Answer 1 · 2018-01-29T08:59:28

$9+10i$

If we have a complex number such that $z=a+bi$ , then $z=r(costheta+isintheta)$ , where:

For $z_1=7+4i$ :
$r=sqrt(7^2+4^2)=sqrt(49+16)=sqrt(65)$
$theta=tan^(-1)(4/7)~~29.7$
$=sqrt(65)(cos(29.7)+isin(29.7))$

For $z_2=2+6i$ :
$r=sqrt(2^2+6^2)=sqrt(4+36)=sqrt(40)$
$theta=tan^(-1)(6/2)~~71.6$
$=sqrt(40)(cos(71.6)+isin(71.6))$

For $z_1+z_2$ :
$z_1+z_2=sqrt(65)cos(29.7)+isqrt(65)sin(29.7)+sqrt(40)cos(71.6)+isqrt(40)sin(71.6)$

$color(white)(z_1+z_2)=sqrt(65)cos(29.7)+sqrt(40)cos(71.6)+i(sqrt(65)sin(29.7)+sqrt(40)sin(71.6))$

$color(white)(z_1+z_2)=8.999470954+9.995734316i$ #

$color(white)(z_1+z_2)~~9+10i$

Proof:
$7+4i+2+6i=(7+2)+i(4+6)=9+10i$

How do you add (7+4i)(7+4i) and (2+6i)(2+6i) in trigonometric form?