How do you add $(- 2 + 5 i) + (- 4 - 2 i)$ $(-2 + 5i) + (-4 - 2i)$ ?

Question

1822 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-27T18:22:01

$- 6 + 3 i$ $-6+3i$

Complex numbers in rectangular form may be added directly, ie you simply add the real and imaginary components individually.

That is, $(a + i b) + (x + i y) = (a + b) + i (x + y)$ $(a+ib)+(x+iy)=(a+b)+i(x+y)$

So in this particular case :

$(- 2 + 5 i) + (- 4 - 2 i) = (- 2 - 4) + i (5 - 2)$ $(-2+5i)+(-4-2i)=(-2-4)+i(5-2)$

$= - 6 + 3 i$ $=-6+3i$ .

How do you add (−2+5i)+(−4−2i)(-2 + 5i) + (-4 - 2i)?