How do you add $(-8+4i)$ and $(-3-3i)$ in trigonometric form?

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Answer 1 · 2016-02-15T09:17:45

$sqrt(122)[cos(tan^-1(1/(-11)))+isin(tan^-1(1/(-11)))]" " "$ OR
$sqrt(122)[cos(174.806^@)+isin(174.806^@)]" " "$

Add $(-8+4i)$ and $(-3-3i)$ to obtain

$-11+i$

Use the formula

$a+bi=sqrt(a^2+b^2)[cos(tan^-1 (b/a))+i*sin(tan^-1 (b/a)]$

Let $a=-11$ and $b=1$

magnitude $" " " "r=sqrt(a^2+b^2)=sqrt((-11)^2+1^2)=sqrt122$

the angle $theta=tan^-1 (1/(-11))=174.806^@$

$sqrt(122)[cos(tan^-1(1/(-11)))+isin(tan^-1(1/(-11)))]" " "$ OR
$sqrt(122)[cos(174.806^@)+isin(174.806^@)]" " "$

have a nice day... from the Philippines..

How do you add (-8+4i)(-8+4i) and (-3-3i)(-3-3i) in trigonometric form?