2012 AIME II Problems/Problem 6

Problem 6

Let $z=a+bi$ be the complex number with $\vert z \vert = 5$ and $b > 0$ such that the distance between $(1+2i)z^3$ and $z^5$ is maximized, and let $z^4 = c+di$. Find $c+d$.


Let's consider the maximization constraint first: we want to maximize the value of $|z^5 - (1+2i)z^3|$ Simplifying, we have

$|z^3| * |z^2 - (1+2i)|$

$=|z|^3 * |z^2 - (1+2i)|$

$=125|z^2 - (1+2i)|$

Thus we only need to maximize the value of $|z^2 - (1+2i)|$.

To maximize this value, we must have that $z^2$ is in the opposite direction of $1+2i$. The unit vector in the complex plane in the desired direction is $\frac{-1}{\sqrt{5}} + \frac{-2}{\sqrt{5}} i$. Furthermore, we know that the magnitude of $z^2$ is $25$, because the magnitude of $z$ is $5$. From this information, we can find that $z^2 = \sqrt{5} (-5 - 10i)$

Squaring, we get $z^4 = 5 (25 - 100 + 100i) = -375 + 500i$. Finally, $c+d = -375 + 500 = \boxed{125}$

Solution 2

WLOG, let $z_{1}=5(\cos{\theta_{1}}+i\sin{\theta_{1}})$ and


This means that



Hence, this means that




Now, common sense tells us that the distance between these two complex numbers is maxed when they both are points satisfying the equation of the line $yi=mx$, or when they are each a $180^{\circ}$ rotation away from each other.

Hence, we must have that $5\theta_{1}=3\theta_{1}+\theta_{2}+180^{\circ}\implies\theta_{1}=\frac{\theta_{2}+180^{\circ}}{2}$

Now, plug this back into $z_{1}^4$(if you want to know why, reread what we want in the problem!)

So now, we have that $z_{1}^4=625(\cos{2\theta_{2}}+i\sin{2\theta_{2}})$

Notice that $\cos\theta_{2}=\frac{1}{\sqrt{5}}$ and $\sin\theta_{2}=\frac{2}{\sqrt{5}}$

Then, we have that $\cos{2\theta_{2}}=\cos^2{\theta_{2}}-\sin^2{\theta_{2}}=-\frac{3}{5}$ and $\sin{2\theta_{2}}=2\sin{\theta_{2}}\cos{\theta_{2}}=\frac{4}{5}$

Finally, plugging back in, we find that $z_{1}^4=625(-\frac{3}{5}+\frac{4i}{5})=-375+500i$


See Also

2012 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AIME Problems and Solutions

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