How do you combine $(- 9 + i) + (6 - 2 i)$ $(-9 + i) + (6-2i)$ ?

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Answer 1 · 2015-11-07T21:32:10

$- 3 - i$ $-3-i$

You may add real and complex parts separately for a complex number in rectangular form.

$therefore(-9+i)+(6-2i)=(-9+6)+i(1-2)$

$=-3-i$

How do you combine (-9 + i) + (6-2i)(−9+i)+(6−2i)(-9 + i) + (6-2i)?