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 $\boxed{289}$ .

1988 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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1988 AHSME Problems/Problem 21

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