+1712 \(\frac{3}{2+3i}\)
Help me, please!
\(tommarvoloriddle \)
+219 \(\frac{3}{2+3i}\)
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
+434 Jeesh, that's the way you thank the person that just helped you? XD If your getting help I would refrain from pink girl.
+1712 What, isn't it a complement? shes good at something- being pink!
+1712 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!
\(tommarvoloriddle\)
+219 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
+1712 I know, I said, "CONFIRMED" so I basically took a look at yours, said, "seems legit" checked mine, made a smiley face emoji: :-)
