Finding a complex number

>>18728211
Just post a picture of your homework.

Don't you remember the formulas to covert real numbers to imaginary and back?

https://www.youtube.com/watch?v=ysVcAYo7UPI&list=PLGR_7q6BJHQGo3lIEUzEWrCLFleDRzvVM

One of these videos should help you

>>18728211
Replace both zs with a+bi.

Make denominator of fraction a real number. You should already know how to do this.

. . . the rest should be trivial.

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>>18728221
Here's a picture, it's exercise 5
I do understand how powers of i work, but I don't see how that's supposed to help me here.

>>18728246
I did that, but you end up with a denominator that's four terms long and simplifies nothing

>>18728246
Here's a write out.
I just end up with garbage, am I doing something wrong?

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>>18728697
Fuck

>>18728702
You should multiply by (A-Bi-2)/(A-Bi-2), not (A+Bi-2)/(A+Bi-2)

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>>18728826
True.
However, I'm not closer to any meaningful answer, I think

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Here you have anon.

>>18729015
I'm afraid I don't get the last three lines.
Why are you keeping the - 2bi outside of the brackets?
Why would the argument of z be pi/2 if a^2 + b^2 - 2a=0?
What do you do between the final step and the one before that?

>>18729015
On your last 3 lines where it says arg(z) = Pi/2, I think it should be z/(z-2), not z. >>18728662

>>18729145
>Why are you keeping the - 2bi outside of the brackets?
>Why would the argument of z be pi/2 if a^2 + b^2 - 2a=0
Do you know what the question is really asking when it's talking about complex arguments? Please study the following references.

https://en.m.wikipedia.org/wiki/Argument_(complex_analysis)

https://www.mathsisfun.com/polar-cartesian-coordinates.html

http://m.wolframalpha.com/input/?i=arctan%28x%29+as+x+approaches+infinity

>What do you do between the final step and the one before that?
He completed the square.

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>>18729145
Hello.

1. I separate the real and imaginary part.
2. You said it yourself: z=ki meaning that if z=w+ki then w=0.
3. Add 1 to both sides. Notice a^2-2a+1=(a-1)^2. Remember that a circumference has equation (x-n)^2+(y-m)^2=r^2, where (n,m) is the center and r the radius, thus we have a circumference centered at (1,0) with radius 1.

>>18729389
I think I get it now, thanks man.