How do you add $(-5-2i)+(7-6i)$ in trigonometric form?

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Answer 1 · 2017-07-15T17:31:01

$(-5-2i) +(7-6i) = 2sqrt17(cos1.33-isin1.33)$

$(-5-2i)+(7-6i) = 2-8i$

This can be written in trigonometric form as:

$a+bi = r(cosvartheta + i sinvartheta)$ where $r=sqrt (a^2+b^2)$ and $vartheta = arctan(y/x)$

$sqrt(2^2+(-8)^2) = 2sqrt17$

$arctan(-8/2) = -1.33$

$therefore 2-8i = 2sqrt17(cos(-1.33) + isin(-1.33))$

Using the properties of the $sin$ and $cos$ functions, this can be rewritten as:

$2-8i=2sqrt17(cos1.33-isin1.33)$

How do you add (-5-2i)+(7-6i)(-5-2i)+(7-6i) in trigonometric form?