2009 AIME I Problem 2

Attempt Problem 2 of the 2009 AIME I 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 2009 AIME I solutions, or check the answer key.

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There is a complex number

z

with imaginary part

164

and a positive integer

n

such that

\frac{z}{z + n} = 4i.

Find

n.

Answer

Write

z = a + 164i

and clear the denominator:

z = 4i(z + n)

Expand

4i(z + n)

and match real parts, then imaginary parts, to find

a

and then

n

Write

z = a + 164i.

Clearing the denominator gives

z = 4i(z + n),

that is,

\begin{aligned} a + 164i &= 4i\,(a + n + 164i) \\ &= -656 + 4(a + n)i. \end{aligned}

Real parts give

a = -656,

and imaginary parts give

164 = 4(a + n),

a + n = 41

and

n = 41 + 656 = 697.

Problem 2 in other years