Question

can anybody help me with this?

Answer 1

<shrugs>
reciprocal of any number, z, is just
\[\Large \frac1z\]
:3

Answer 2

and how do i Solve this problem? can you please explain?

Answer 3

I can imagine your frustration.... but let's start with what I said... for all intents and purposes, the reciprocal of 3-2i is just
\[\Large \frac1{3-2i}\] yes? :)

Answer 4

yes

Answer 5

Now, the folks behind this complex number system DO NOT LIKE the imaginary unit being in the denominator... (lol haters)
So somehow, we have to express this
\[\Large \frac1{3-2\color{red}i}\] such that that imaginary unit \(\large \color{red}i\) does not appear in the denominator... any ideas how to do that? ^_^

Answer 6

Incidentally, have you heard of the concept of 'conjugates' when it comes to complex numbers?

Answer 7

yeah, conjugate, like the opposite?

Answer 8

Okay, so the trick here is to multiply both the numerator and the denominator by the conjugate of 3 - 2i

Bearing in mind that the product of a complex number and its conjugate is always a real number....

Having said that, what is the conjugate of 3 - 2i ?

Answer 9

[Direct Hint]
For any complex number \[\Large z = a+ b\color{blue}i\]
Its conjugate is always given by \[\Large \bar z = a - b\color{blue}i\]


In other words, just change the sign of the imaginary part ^_^

Answer 10

3+2i right?

Answer 11

That is correct :P
So let's do this: basically multiplying everything by 1, but a rather 'special' variant of 1.
\[\Large \frac{1}{3 - 2\color{blue}i}\cdot \frac{3+2\color{blue}i}{3+2\color{blue}i}\]
Please simplify :)

Answer 12

[Subtle Hint]
\[\Large (x+y)(x-y) = x^2 -y^2\]

Answer 13

\[\frac{ 1+3+2i }{ 3^{2} - (2i)^{2} }\]

Answer 14

is this correct?

Answer 15

Nope :P
Why did you add the numerators? We're multiplying fractions, lol
\[\Large \frac{\alpha}{\beta}\cdot \frac{\gamma}{\delta}= \frac{\alpha\gamma}{\beta\delta}\]

Answer 16

Though the denominator, you got correctly... fix the numerator... come on, what is 1 times anything? XD

Answer 17

\[\frac{ 1 \times 3 + 2i }{ (3)^{2} - (2i)^{2} }\]

Answer 18

like this?

Answer 19

Close enough... I'd prefer you have it as
\[\Large \frac{1(3+2\color{blue}i)}{(3)^2 - (2\color{blue}i)^2}\]

Answer 20

Now the numerator is just... 3 + 2i, right?

Answer 21

yes

Answer 22

Because 1 times any number is that number...
\[\Large \frac{3+2\color{blue}i}{3^2 -(2\color{blue}i)^2}\]
Now, the finale, I'm stepping back and letting you finish it off... make your country proud....
<salutes>

:D

Answer 23

Simplify the denominator...

Answer 24

...psst... your audience does not like to be kept waiting, lol...
What is \[\Large 3^2 = \color{red}?\]

Answer 25

9

Answer 26

I love the way you are helping me!!!

Answer 27

Say that when we're finished ... lol
what is
\[\Large (2\color{blue}i)^2=\color{red}?\]

Answer 28

4iˆ2

Answer 29

Yes... and we know (only too well) that
\[\Large \color{blue}i^2 = \color{red}?\]
Come on, it's like the first thing you learn when you start complex numbers...

Answer 30

-1

Answer 31

so 9 - (-4)

Answer 32

That's right... so...
\[\Large (2\color{blue}i)^2 = 4\color{blue}i^2 = \color{red}?\]

Answer 33

Oh, lol, you're way ahead of me XD

Answer 34

Well with your professional help :)

Answer 35

Okay, great, I love initiative, (especially when it's correct)
Let me just arrange the masterpiece for you :3
\[\Large = \frac{3+2\color{blue}i}{9-(-4)}\]

time for the grand finale... finish it, maestro :3

Answer 36

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

Answer 37

Et voila ^_^

Answer 38

THANK YOU SO MUUUCH!!! YOU ARE AMAZING. I wish you were my math teacher!

Answer 39

Come now, let's not deprive your actual teacher of credit just yet :P

Answer 40

Now, to wrap it up, I'd like to generalize it for you...
\[\Large \frac1z = \frac{\bar z}{|z|^2}\]

If you don't like it like that, then
\[\Large \frac{1}{a+b\color{blue}i}= \frac{a-b\color{blue}i}{a^2+b^2}\]
This works every time :P

Answer 41

Just so happened that in this case
a = 3
and
b = -2

Cheers to entertainment LOL ^_^

Answer 42

THANK YOU SO MUCH!

Answer 43

No problem :)