Question

Perform the indicated operation and express your answer in standard form a+bi

\[\frac{4}{1-2i}-\frac{2}{3+i}\]

Answer 1

\[\implies \frac{4(3+i)-2(1-2i)}{(1-2i)(3+i).}\]U can do it now:)

Answer 2

Yeah but I am stuck expressing it as a+bi, I am here:

\[\frac{10+8i}{5-5i}\]

Answer 3

So what you must do is multiply by the complex conjugate of the denominator! (5+5i)/(5+5i) multiply by that!

Answer 4

\[\frac{10+8i}{5-5i} \times \frac{5+5i}{5+5i}.\]

Answer 5

Okay...

Answer 6

\[\frac{4}{1-2i}-\frac{2}{3+i}\]\[=\frac{4(1+2i)}{(1-2i)(1+2i)}-\frac{2(3-i)}{(3+i)(3-i)}\]\[=\frac{4+8i}{1-(-4)}-\frac{6-2i}{9-(-1)}\]\[=\frac{4+8i}{5}-\frac{6-2i}{10}\]\[=\frac{2(4+8i)}{5(2)}-\frac{6-2i}{10}\]\[=\frac{8+16i-6+2i}{10}\]\[=\frac{2+18i}{10}\]\[=\frac{2}{10}+\frac{18i}{10}\]\[=\frac{1}{5}+\frac{9i}{5}\]

Answer 7

Okay, thanks so much everyone!!! I got the same answer as @RolyPoly...

Answer 8

^_^

Answer 9

You can multiply by the complex conjugate before or after combining the fractions, it doesn't matter the order.

Answer 10

I think I'll do it after @Kainui but thanks so much for the tip :) !