Find all complex numbers $z$ such that
\[
|z|^2-2\bar z+iz=2i.
\]
|z|^2-2z*+iz = 2i
letting z = x+yi,
x^2+y^2 - 2(x-yi) + i(x+yi) = 2i
x^2+y^2-2x+2yi+ix-y-2i = 0
(x^2+y^2-2x-y)+(x+2y-2)i = 0
so, now just solve
x^2+y^2-2x-y=0
x+2y = 2
x = 2-2y
(2-2y)^2+y^2-2(2-2y)-y=0
5y^2-5y = 0
y = 0 or 1
so, x = 2 or 0
z = 2, i
I am confused on how z = 2 or i
Hi Pumple, it is z = 2, i because z is a complex number so the y is 1 so it is 1i, it is just that the 1 is invisible, but its there
To find all complex numbers $z$ that satisfy the given equation, we need to simplify the equation and express it in terms of $z$. Let's start by expanding the conjugate expression $\bar z$.
The equation is:
\[|z|^2 - 2\bar{z} + iz = 2i.\]
Since $z$ is a complex number, we can write $z$ in terms of its real and imaginary parts: $z = x + yi,$ where $x$ and $y$ are real numbers. Using this notation, the conjugate of $z$ is $\bar{z} = x - yi.$
Now, let's rewrite the equation in terms of $x$ and $y:$
\[|x + yi|^2 - 2(x - yi) + i(x + yi) = 2i.\]
Simplifying each part, we have:
\[(x+yi)(x-yi) - 2(x - yi) + i(x + yi) = 2i.\]
\[x^2 - xyi + xyi - (yi)^2 - 2x + 2yi + xi + yi^2 = 2i.\]
\[x^2 + y^2 + (3y - x)i = 2i.\]
Since the equation should hold true for all complex numbers $z,$ the real and imaginary parts on both sides of the equation must be equal. Thus, we have two equations:
Real part: $x^2 + y^2 = 0.$
Imaginary part: $3y - x = 2.$
From the first equation, we observe that $x^2 + y^2 = 0$ if and only if $x = 0$ and $y = 0.$
Substituting $x = 0$ into the second equation, we get $3y = 2,$ which gives $y = \frac{2}{3}.$
Therefore, the only complex number $z$ that satisfies the equation is $z = \frac{2}{3}i.$