HMMT February

Let $|a-z| /|b-z|=k$. We wish to determine the minimum and maximum value of $k$. Squaring and expansion give: $|a-z|^{2} =|b-z|^{2} \cdot k^{2} |a|^{2}-2 a \cdot z+1 =\left(|b|^{2}-2 b \cdot z+1\right) k^{2} |a|^{2}+1-\left(|b|^{2}+1\right) k^{2} =2\left(a-b k^{2}\right) \cdot z$ where $\cdot$ is a dot product of complex numbers, i.e., the dot product of vectors corresponding to the complex numbers in the complex plane. Now, since $z$ has modulus 1 but can assume any direction, the only constraint on the value of $k$ is $\left.|| a\right|^{2}+1-\left(|b|^{2}+1\right) k^{2}|\leq 2| a-b k^{2} \mid$. Squaring again and completing the square, the inequality reduces to: $\left(|a|^{2}-1\right)^{2}+\left(|b|^{2}-1\right)^{2} k^{4}+2\left(4 a \cdot b-\left(|a|^{2}+1\right)\left(|b|^{2}+1\right)\right) k^{2} \leq 0 \left(\left(|a|^{2}-1\right)-\left(|b|^{2}-1\right) k^{2}\right)^{2}-4|a-b|^{2} k^{2} \leq 0 \left|\left(|a|^{2}-1\right)-\left(|b|^{2}-1\right) k^{2}\right| \leq 2|a-b| k$. At this stage all the relevant expressions are constant real numbers. Denote, for simplicity, $A=|a|^{2}-1, B=|b|^{2}-1$, and $C=|a-b|$. Then, we are looking for $k$ such that $\left|A-B k^{2}\right| \leq 2 C k$. If $B=0$, then $k \geq\left|\frac{A}{2 C}\right|$, so the minimum value is $\left|\frac{A}{2 C}\right|$ and the maximum value is $+\infty$. Otherwise, consider $C^{2}+A B =\left(|a|^{2}-2 a \cdot b+|b|^{2}\right)+\left(|a|^{2}-1\right)\left(|b|^{2}-1\right) =|a b|^{2}-2 a \cdot b+1 =|\bar{a} b|^{2}-2 \Re(\bar{a} b)+1 =|\bar{a} b-1|^{2}$. So let $D=|\bar{a} b-1|=\sqrt{C^{2}+A B}$. We may assume $B>0$ (the another case is analogous: just substitute $A, B$ with $-A,-B)$. Then, $k$ is determined by the following inequalities: $B k^{2}+2 C k-A \geq 0$ and $B k^{2}-2 C k-A \leq 0$. The first inequality gives $k \leq \frac{-C-D}{B}$ or $k \geq \frac{-C+D}{B}$, and the second gives $\frac{C-D}{B} \leq k \leq \frac{C+D}{B}$. Combining, this gives $\left|\frac{C-D}{B}\right| \leq k \leq\left|\frac{C+D}{B}\right|$, as claimed. To summarize the general answer, let $A=|a|^{2}-1, B=|b|^{2}-1, C=|a-b|, D=|\bar{a} b-1|$. Then, if $|b|=1$, min is $\left|\frac{A}{2 C}\right|$ and max is $+\infty$; otherwise, $\min$ is $\left|\frac{C-D}{B}\right|$ and $\max$ is $\left|\frac{C+D}{B}\right|$. In the special case $a=\sqrt{17}$ and $b=\sqrt{19} i$, we have $A=16, B=18, C=|\sqrt{17}-\sqrt{19} i|=\sqrt{36}=6$, and $D=\sqrt{17 \cdot 19+1}=18$. Thus the answer is $\frac{C+D}{B}=\frac{6+18}{18}=\frac{4}{3}$.