# Difference between revisions of "1988 AHSME Problems/Problem 21"

## Problem

The complex number $z$ satisfies $z + |z| = 2 + 8i$. What is $|z|^{2}$? Note: if $z = a + bi$, then $|z| = \sqrt{a^{2} + b^{2}}$.

$\textbf{(A)}\ 68\qquad \textbf{(B)}\ 100\qquad \textbf{(C)}\ 169\qquad \textbf{(D)}\ 208\qquad \textbf{(E)}\ 289$

## Solution

Let the complex number $z$ equal $a+bi$. Then the preceding equation can be expressed as $$a+bi+\sqrt{a^2+b^2} = 2+8i$$ Because $a$ and $b$ must both be real numbers, we immediately have that $bi = 8i$, giving $b = 8$. Plugging this in back to our equation gives us $a+\sqrt{a^2+64} = 2$. Rearranging this into $2-a = \sqrt{a^2+64}$, we can square each side of the equation resulting in $$4-4a+a^2 = a^2+64$$ Further simplification will yield $60 = -4a$ meaning that $-15 = a$. Knowing both $a$ and $b$, we can plug them in into $a^2+b^2$. Our final answer is $\boxed{289}$.