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April 10th, 2016, 12:29 AM   #1
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Complex numbers equation

**Edit: The original equation is: izz'=z-z' (z' is the conjugate, I don't know how to type it here)

the final expression in my equation is:

isin(90)=2*isin(theta).


I know that 1 working way is to do:

*sin(90)=1, so you get:

i=2*isin(theta)

>> 1=2sin(theta) > > sin(theta)=(1/2) >> theta=30

But how do I know from here that the equation has 2 solutions? and what is the second solution?

I also tried to do (which didn't work) the following:

2sin(theta)=90+360k >> sin(theta)=45+180k < and you can see from this one that it has 2 solutions, but they are wrong

Last edited by noobinmath; April 10th, 2016 at 01:18 AM.
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April 10th, 2016, 04:01 AM   #2
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$\displaystyle iz\bar{z}=z-\bar{z}$

$\displaystyle i(a^2+b^2)=(a+bi)-(a-bi)=2bi$

$\displaystyle a^2+b^2=2b,\quad a=0,b=2;\quad a=1,b=1$

I don't see any other solutions (I'm assuming you want to find $z$ in rectangular form; please pardon me if that's not the case).
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April 11th, 2016, 01:36 AM   #3
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Thank you!
using the rectengular forum as you did, I can see why it has 2 solutions,

but if I needed to solve it in the polar form, how can I tell how many solutions does it have? because, for example when I have z^5=<expression> I know it should have 5 solutions, but when the equation is a mix of z and z conjugate, how do you know that, for example in my equation that it should have 2 solutions?
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April 11th, 2016, 08:07 PM   #4
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That would depend on the equation.
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