• March 9th 2010, 08:38 AM
Mhockey02
Find (2+3i)-(-5+8i)

also

Find (5-9i)(2+3i)
• March 9th 2010, 09:11 AM
e^(i*pi)
Quote:

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 $i^2 = -1$
• March 9th 2010, 09:31 AM
Mhockey02
So the answer for the first one is -4 and the second one is 9?
• March 9th 2010, 09:56 AM
e^(i*pi)
Quote:

Originally Posted by Mhockey02
So the answer for the first one is -4 and the second one is 9?

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