Question 1085367
<font color="black" face="times" size="3">I'm assuming the initial equation is *[Tex \LARGE |z|^2-2\bar{z}+iz=2i]


If so, then let {{{z = a+bi}}} where *[Tex \Large a,b \in \mathbb{R}] (a,b are real numbers)


The complex conjugate of {{{z}}} is *[Tex \LARGE \bar{z} = a-bi]


Using the pythagorean theorem, and a visual representation of {{{z}}}, we can say *[Tex \Large |z|^2 = a^2+b^2]


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Using those equations, we can make a bit of substitutions and rearrangements to get the following


*[Tex \LARGE |z|^2-2\bar{z}+iz=2i]


*[Tex \LARGE a^2+b^2-2(a-bi)+i(a+bi)=2i]


*[Tex \LARGE a^2+b^2-2a+2bi+ai+bi^2=2i]


*[Tex \LARGE a^2+b^2-2a+2bi+ai+b(-1)=2i]


*[Tex \LARGE a^2+b^2-2a+2bi+ai-b=2i]


*[Tex \LARGE (a^2+b^2-2a-b)+(2bi+ai)=2i]


*[Tex \LARGE (a^2+b^2-2a-b)+(2b+a)*i=0+2*i]


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From that last equation, in the section above, we can equate the real and imaginary parts to form these two equations


{{{a^2+b^2-2a-b = 0}}}


{{{2b+a = 2}}}


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Solve {{{2b+a = 2}}} for 'a'


{{{2b+a = 2}}}


{{{2b+a-2b = 2-2b}}}


{{{a = 2-2b}}}


Now plug this into {{{a^2+b^2-2a-b = 0}}} and solve for b


{{{a^2+b^2-2a-b = 0}}}


{{{(2-2b)^2+b^2-2(2-2b)-b = 0}}}


{{{4-8b+4b^2+b^2-4+4b-b = 0}}}


{{{(4b^2+b^2)+(-8b+4b-b)+(4-4) = 0}}}


{{{5b^2 - 5b = 0}}}


{{{5b(b - 1) = 0}}}


{{{5b=0}}} or {{{b - 1 = 0}}}


{{{b=0}}} or {{{b = 1}}}


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If {{{b=0}}}, then 


{{{a = 2-2b}}}


{{{a = 2-2*0}}}


{{{a = 2}}}


Therefore one solution is {{{z = a+bi = 2+0i = 2}}} which is purely a real number.


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If {{{b = 1}}}, then


{{{a = 2-2b}}}


{{{a = 2-2*1}}}


{{{a = 0}}}


The other solution is {{{z = a+bi = 0+1i = i}}} which is a <a href="http://mathworld.wolfram.com/PurelyImaginaryNumber.html">purely imaginary number</a>


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Summary:


The two solutions are {{{z = 2+0i}}} (which is the same as {{{z = 2}}}) and {{{z = 0+1i}}} (which is the same as {{{z = i}}})</font>