Imaginary numbers (simplify a + bi)

[foahchon]

I have a problem involving imaginary numbers that I can't quite figure out how to solve (book does not provide a model problem). The instructions say to simplify this:

1/(1 - 3i) - 3/(4 + 2i)

The back of the book says the answer is -2/5 + (9/10)i, but no solution I can figure comes anywhere near this.

Can someone help me figure this out? Thanks.

 

[soroban]

Hello, foahchon!

I too get a different answer . . . the book must be wrong.

\(\displaystyle \L\frac{1}{1\,-\,3i} \, -\,\frac{3}{4\,+\,2i}\)

We must rationalize those denominators.

The first fraction is: \(\displaystyle \L\:\frac{1}{1\,-\,3i}\,\cdot\,\frac{1\,+\,3i}{1\,+\,3i} \:=\:\frac{1\,+\,3i}{1\,-\,9i^2} \:=\:\frac{1\,+\,3i}{10}\)

The second fraction is: \(\displaystyle \L\:\frac{3}{4\,+\,2i}\,\cdot\,\frac{4\,-\,2i}{4\,-\,2i} \:=\:\frac{3(4\,-\,2i)}{16 - 4i^2} \:=\:\frac{12\,-\,6i}{20} \:=\:\frac{6\,-\,3i}{10}\)


The problem becomes: \(\displaystyle \L\:\frac{1\,+\,3i}{10} \,-\,\frac{6\,-\,3i}{10} \;=\;\frac{(1\,+\,3i)\,-\,(6\,-\,3i)}{10} \;=\;\frac{1\,+\,3i\,-\,6\,+\,3i}{10}\)

. . \(\displaystyle \L= \;\frac{-5\,+\,6i}{10}\;=\;\frac{-5}{10}\,+\,\frac{6i}{10} \;=\;\fbox{-\frac{1}{2}\,+\,\frac{3}{5}i}\)

 

[foahchon]

That's quite a bit different from what I tried, I'll give that a shot. Thank you for your help.