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avatar+1675 

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

 

Help me, please!

 

\(tommarvoloriddle \)

 Jul 10, 2019
 #1
avatar+219 
+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

 Jul 10, 2019
 #5
avatar+105654 
+4

Very nice presentation EmeraldWonder. laugh

Melody  Jul 10, 2019
 #6
avatar+219 
+6

Thank you very much, Melody ;P

EmeraldWonder  Jul 10, 2019
 #7
avatar+1675 
+1

Nice, Pink Girl!

tommarvoloriddle  Jul 11, 2019
 #8
avatar+429 
+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
 #9
avatar+1675 
+1

What, isn't it a complement? shes good at something- being pink!

tommarvoloriddle  Jul 11, 2019
 #10
avatar+429 
+1

Oh wow, that's just rude XD

HiylinLink  Jul 12, 2019
 #11
avatar+1675 
+1

What? She does have a "pinky" avatar...

tommarvoloriddle  Jul 12, 2019
 #2
avatar+1675 
+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!

\(tommarvoloriddle\)

.
 Jul 10, 2019
 #3
avatar+219 
+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
avatar+1675 
+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

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