+219 32+3i
To remove 2+3i from the denominator, multiply the numerator and denominator by its complex conjugate
3(2−3i)(2+3i)(2−3i)
Solve the denominator seperately
(2+3i)(2−3i)
22+32
13
Now you have
3(2−3i)13
Expand 3(2−3i) by distributing
3(2−3i)
6−9i
Now you would have
6−9i13=613−913i
Your answer would be
613−913i
Hope this Helps ;P
+448 Jeesh, that's the way you thank the person that just helped you? XD If your getting help I would refrain from pink girl.
+1713 What, isn't it a complement? shes good at something- being pink!
+1713 Thank you, Emerald Wonder (pink girl)! I confirmed my answer with yours- and I got 6−9i13
I think that is close enough.
PEACE!
tommarvoloriddle
+219 If you look at the work I presented to you, you would see that 6−9i13 is the answer I got I just simplified it as 613−9i13 which are basically the same thing.
But your welcome, happy to help ;P
+1713 I know, I said, "CONFIRMED" so I basically took a look at yours, said, "seems legit" checked mine, made a smiley face emoji: :-)
