Let z = x+yi
Then we have
x^2+y^2-2x+2yi+xi-y=2i
That means we have
x^2+y^2-2x-y = 0
2y+x-2 = 0
Thus, x = 2-2y
and so
4-8y+4y^2+y^2-4+4y-y = 0
5y^2-5y = 0
5y(y-1) = 0
so,
y=0 x=2
or
y=1 x=0
Or,
z=2 or z=i
I'll let you verify that both of those z values satisfy the original equation.
Hey! So, I've been working on this problem for awhile but I'm currently a bit dumbfounded on how to solve it. I think this question has been answered before but I still don't quite understand how it works or how to explain it.
The problem is as follows:
|z|^2 -2(complex conjugate of z) + iz= 2i
It'd really love some help with this problem. So far, I'm lost, but I don't know why that is the case or how I would be able to solve this beyond guessing and checking. The question is looking for complex answers, and I tried substituting z for (a + bi), but that wasn't really very helpful, either due to my lack of understanding or getting lost somewhere.
Any and all help would be greatly appreciated!
2 answers
Thanks for the response Steve, but this doesn't quite clarify things for me. The step I keep getting lost on is this one:
"x^2+y^2-2x-y = 0
2y+x-2 = 0"
How did you get to this point? What happened to the 2yi + xi?
"x^2+y^2-2x-y = 0
2y+x-2 = 0"
How did you get to this point? What happened to the 2yi + xi?