Question

Simplify..

Answer 1

\[\frac{ 13 }{ -3+2i }\]

Answer 2

@waterineyes @Jhannybean

Answer 3

If complex number is in denominator, we use rationalization Method to make our denominator rational..

Answer 4

Have you studied conjugate of complex number?

Answer 5

When you have a complex number in the denominator, in order to get rid of it you want to treat it a a function.

When we have a square root in the denominator of a function, you have learned that in order to get rid of it, yu have t multiply by the conjugate right?

this brings us back to our known identity \(i=\sqrt{-1}\) ,

Answer 6

Sorry this thing was being dumb but yes Water, I have.

Answer 7

So we will have \[\frac{13}{2i-3} \cdot \frac{(2i+3)}{(2i+3)} = \frac{13(3+2i)}{(2i-3)(2i+3)}\]

Answer 8

Now you can simplify the denominator like we had done in your previous problem by expanding the function and replacing \(i^2 \iff -1\)

Answer 9

Do I simplify?

Answer 10

If \(a+ib\) is some complex number, then on changing the sign of imaginary part of this complex number, you get conjugate of this complex number:

Imaginary part has sign of \(+\), so its conjuage will be : \(a-ib\)..

Answer 11

I got \[\frac{ 26+2i }{ 4i ^{2}-9 }\]

Answer 12

Sorry, OS went down for a bit.

Answer 13

Yeah its ok.

Answer 14

And so substitute (-1) in for the i in the denominator, what do you get?

Answer 15

-4^2-9

Answer 16

So its 4^2-9

Answer 17

I'm writing a negative where there doesn't need to be, Hmmm

What is 4(-1) -9 = ?

Answer 18

-13?

Answer 19

I'm not quite sure why you are squaring the 4. Can you explain that to me?

Answer 20

Yes, -13.

Answer 21

Before we move on, i'd like your reasoning on that :)

Answer 22

I had actually the I because when you said "So we will have" at the very top, there were two i's..

Answer 23

*I had actually squared the i because...

Answer 24

Oh yes, we have 2 i's there but foiling the bottom out gave us :\[(2i-3)(2i+3)\]\[=4i^2 +6i - 6i -9\]\[=4\color{red}{i^2} -9 ~~ ~~~~~~~~~~~~~~~~~~~\color{red}{i^2 \iff -1}\]\[=4(-1)-9\]\[=-4-9\][=-13\]

Answer 25

\[=-13\]

Answer 26

OH ok.

Answer 27

This is exactly how you are going to approach every problem that you come across ike this.

Answer 28

ok

Answer 29

But, moving on, from the numerator we had \[\frac{13(3+2i)}{-13}\]We can reduce this and we will get -1 multiplied by the whole thing.\[=-(3+2i)\]

Answer 30

Oh ok. Thanks!

Answer 31

You had tried multiplying in the 13 at first, I had seen that, but I believe you multiplied it in wrong as you had \((26 +2i)\)

Answer 32

No problem :) good luck.

Answer 33

Yeah that's what happened! But thank so much!

Answer 34

not a problem :)

Answer 35

to be very specific, that is not called conjugate that bean has used above..

Answer 36

This is true that you will get answer by that also, but that is not said, "multiplying and dividing by its conjugate"..

Answer 37

What do you call them? inverse of the complex number?

Answer 38

Like I said, Conjugate we get by reversing the sign of Imaginary part and not real part..

Answer 39

I see, it's simpply multiplying by the inverse of the complex number.