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 April 10th, 2016, 12:29 AM #1 Senior Member   Joined: Oct 2014 From: Complex Field Posts: 119 Thanks: 4 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. April 10th, 2016, 04:01 AM #2 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond $\displaystyle iz\bar{z}=z-\bar{z}$ $\displaystyle i(a^2+b^2)=(a+bi)-(a-bi)=2bi$ $\displaystyle a^2+b^2=2b,\quad a=0,b=2;\quad a=1,b=1$ 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). Thanks from noobinmath April 11th, 2016, 01:36 AM #3 Senior Member   Joined: Oct 2014 From: Complex Field Posts: 119 Thanks: 4 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= 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? April 11th, 2016, 08:07 PM #4 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond That would depend on the equation. Tags complex, equation, numbers Search tags for this page

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