The complex number z satisfies z+ z=2+8i. What is z^2? Note: if z=a+bi, then z=√a^2+b^2.

1988 AMC 12 Problem 21

Attempt Problem 21 of the 1988 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1988 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

The complex number

z

satisfies

z+\lvert z\rvert=2+8i.

What is

\lvert z\rvert^2?

Note: if

z=a+bi,

then

\lvert z\rvert=\sqrt{a^2+b^2}.

68

100

169

208

289

Write

z=a+bi

and compare imaginary parts

Use the real part to relate

a

\sqrt{a^2+b^2}

Writing

z=a+bi

gives

b=8

and

a+\sqrt{a^2+64}=2.

Thus

\sqrt{a^2+64}=2-a.

Squaring yields

a=-15.

Therefore

\lvert z\rvert^2=a^2+b^2=225+64=289.

Thus the correct answer is E.