We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

#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