#1**+1 **

the two solutions are

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

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

bigbrotheprodude Feb 27, 2019

#5**0 **

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

Guest Feb 27, 2019

#7**+1 **

Find all complex numbers such that

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

**Formula:**

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

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

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

heureka Feb 27, 2019