#1**+7 **

\(\frac{3}{2+3i}\)

To remove \(2+3i\) from the denominator, multiply the numerator and denominator by its complex conjugate

\(\frac{3(2-3i)}{(2+3i)(2-3i)}\)

Solve the denominator seperately

\((2+3i)(2-3i)\)

\({2}^{2}+{3}^{2}\)

\(13\)

Now you have

\(\frac{3(2-3i)}{13}\)

Expand \(3(2-3i)\) by distributing

\(3(2-3i)\)

\(6-9i\)

Now you would have

\(\frac{6-9i}{13}=\frac{6}{13}-\frac{9}{13}i\)

Your answer would be

\(\frac{6}{13}-\frac{9}{13}i\)

Hope this Helps ;P

EmeraldWonder Jul 10, 2019

#8**+2 **

Jeesh, that's the way you thank the person that just helped you? XD If your getting help I would refrain from pink girl.

HiylinLink
Jul 11, 2019

#2**+5 **

Thank you, Emerald Wonder (pink girl)! I confirmed my answer with yours- and I got \(\frac{6-9i}{13}\)

I think that is close enough.

PEACE!

\(tommarvoloriddle\)

.tommarvoloriddle Jul 10, 2019

#3**+6 **

If you look at the work I presented to you, you would see that \(\frac{6-9i}{13}\) is the answer I got I just simplified it as \(\frac{6}{13}-\frac{9i}{13}\) which are basically the same thing.

But your welcome, happy to help ;P

EmeraldWonder
Jul 10, 2019

#4**+1 **

I know, I said, "CONFIRMED" so I basically took a look at yours, said, "seems legit" checked mine, made a smiley face emoji: :-)

tommarvoloriddle
Jul 10, 2019