Write your answer in a + bi form

**Mhockey02** · March 9th 2010, 07:38 AM

Find (2+3i)-(-5+8i)

also

Find (5-9i)(2+3i)

**e^(i*pi)** · March 9th 2010, 08:11 AM

Originally Posted by Mhockey02

Find (2+3i)-(-5+8i)

also

Find (5-9i)(2+3i)

When adding or subtracting complex numbers add/subtract the real component and the imaginary component separately

$(a+bi)-(c+di) = (a-c) + (b-d)i$

When multiplying treat it the same as you would with any polynomial. Just recall that $i^2 = -1$

**Mhockey02** · March 9th 2010, 08:31 AM

So the answer for the first one is -4 and the second one is 9?

**e^(i*pi)** · March 9th 2010, 08:56 AM

Originally Posted by Mhockey02

So the answer for the first one is -4 and the second one is 9?

No, there are no imaginary terms in your answers.

For your first one: $(2+3i)-(-5+8i) = (2-(-5)) + (3-8)i = 7-5i$

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