How do you add $(7+5i)+(2-4i)$ in trigonometric form?

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Answer 1 · 2018-06-13T13:18:36

$sqrt74cos(0.6202)+sqrt20cos(5.1760)+i(sqrt74sin(0.6202)+sqrt20sin(5.1760))$
$9+i$

$z=a+bi=r(costheta+isintheta)$

$r=sqrt(a^2+b^2)$
$theta=tan^-1(b/a)$

$r_1(cos(theta_1)+isin(theta_2))+r_2(cos(theta_2)+isin(theta_2))=r_1cos(theta_1)+r_2cos(theta_2)+i(r_1sin(theta_1)+r_2sin(theta_2))$

$r_1=sqrt(7^2+5^2)=sqrt(49+25)=sqrt74$
$r_2=sqrt(2^2+4^2)=sqrt(4+16)=sqrt20$

$theta_1=tan^-1(5/7)~~0.6202^c$
$theta_2=tan^-1(-2)~~-1.1071^c$

However, as $2-4i$ is in quadrant 4, we need to add $2pi$ to get a positive angle variant.

$theta_2=2pi+tan^-1(-2)~~5.1760^c$

$sqrt74cos(0.6202)+sqrt20cos(5.1760)+i(sqrt74sin(0.6202)+sqrt20sin(5.1760))$
$7+2+i(5-4)$
$9+i$

Proof:

$7+5i+2-4$
$(7+2)+i(5-4)$
$9+i$

How do you add (7+5i)+(2-4i)(7+5i)+(2-4i) in trigonometric form?