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# Help!

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$$\frac{3}{2+3i}$$

$$tommarvoloriddle$$

Jul 10, 2019

#1
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$$\frac{3}{2+3i}$$

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

$$\frac{6-9i}{13}=\frac{6}{13}-\frac{9}{13}i$$

$$\frac{6}{13}-\frac{9}{13}i$$

Hope this Helps ;P

Jul 10, 2019
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Very nice presentation EmeraldWonder.

Melody  Jul 10, 2019
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Thank you very much, Melody ;P

EmeraldWonder  Jul 10, 2019
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Nice, Pink Girl!

tommarvoloriddle  Jul 11, 2019
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Jeesh, that's the way you thank the person that just helped you? XD If your getting help I would refrain from pink girl.

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

tommarvoloriddle  Jul 11, 2019
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Oh wow, that's just rude XD

#11
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What? She does have a "pinky" avatar...

tommarvoloriddle  Jul 12, 2019
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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
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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
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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