I am having problems with the following :
Z +|Z|^2 = 10-2i (the solution must be in the form a+bi)
I have gotten to :
a +bi + a^2 +b^2 = 10 -2i
and cannot think how to solve it
Any help would be appreciated.
Jeff
As you were told, compare the real and imaginary parts.
You have $\displaystyle \displaystyle \begin{align*} a + a^2 + b^2 + b\,i = 10 - 2i \end{align*}$, so $\displaystyle \displaystyle \begin{align*} a + a^2 + b^2 = 10 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} b = -2 \end{align*}$. Solve for $\displaystyle \displaystyle \begin{align*} a \end{align*}$.
two complex numbers:
a+bi and c+di are equal if and only if: a = c, and b = d.
so if a^{2}+b^{2}+ a + bi = 10 - 2i, then:
a^{2} + b^{2} + a = 10, and b = -2.
since b = -2, a^{2} + b^{2} + a = a^{2} + 4 + a.
since we know this already equals 10, we get:
a^{2} + a - 6 = 0
and a^{2} + a - 6 = (a - 2)(a + 3), so a = 2, or -3.
as a final check, we verify that:
z = 2 - 2i
z = -3 - 2i both satisfy:
z + |z|^{2} = 10 - 2i
2 - 2i + |2 - 2i|^{2} = 2 - 2i + 4 + 4 = (2 + 4 + 4) - 2i = 10 - 2i
-3 - 2i + |-3 - 2i|^{2} = -3 - 2i + 9 + 4 = (-3 + 9 + 4) - 2i = 10 - 2i (checked).