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# how do you solve this?

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Find all complex numbers  such that
$$|z|^2-2\bar z+iz=2i.$$

Feb 27, 2019

#1
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the two solutions are

z=2+0i( which is the same as z=2)

z=0+1i( which is the same as z=1

Feb 27, 2019
#2
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Whoa that's hard, what grade are you in whoever has this complex number problem?

Guest Feb 27, 2019
#3
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college

Feb 27, 2019
#4
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thx,

if you need any help just ask

Feb 27, 2019
#5
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This is the person who posted the question responding:

I'm in 7th grade but I take an Algebra A class. I haven't done anything like this before

Feb 27, 2019
#6
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oh okay

i see where did you get it from

Feb 27, 2019
#7
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Find all complex numbers  such that

$$\mathbf{|z|^2-2\bar z+iz=2i}$$

Formula:

$$z=a+bi \\ \bar z = a-bi \\ |z|^2 = a^2+b^2$$

$$\begin{array}{|rcll|} \hline |z|^2-2\bar z+iz &=& 2i \\ a^2+b^2-2(a-bi )+ i(a+bi)&=& 2i \\ a^2+b^2-2a+2bi + ia+bi^2 &=& 2i \quad | \quad i^2=-1 \\ a^2+b^2-2a+2bi + ia -b &=& 2i \\ a^2+b^2-2a-b +2bi + ia &=& 2i \\ \underbrace{(a^2+b^2-2a-b)}_{=0} + \underbrace{(2b+a)}_{=2}i &=& 2i \\\\ 2b+a &=& 2 \\ 2b &=& 2-a \\ \mathbf{b} & \mathbf{=}& \mathbf{\dfrac{2-a}{2}} \\ \hline \end{array}$$

$$\mathbf{z = a + \left(\dfrac{2-a}{2}\right)i}$$

Feb 27, 2019