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$u$ and $v$ are two complex numbers such that $\left|u^2+uv-2v^2-3i\left(u+2v+2\right)+8\right|=44$ and $\left|u+2v\right|=6$. Find maximum value of $T=\left|u-2i\right|^2+2\left|v+i\right|^2$.

I have tried and get some interesting things, but haven't known how to do next, can you help me?

We can easily have $\left|\left(u+2v\right)\left(u-v\right)-3i\left(u+2v\right)-6i+8\right|=44$. With $u+2v=x$ and $u-v-3i=y$, we have $|xy-6i+8|=44$ and $|x|=6$. Interestingly, $v=\frac{x-y}{3}-i$ and $u=\frac{x+2y}{3}+2i$. Now we only need to find max of $|x+2y|^2+2|x-y|^2$.

user628755
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Substituting $\alpha = 6i-8$, $z = xy - \alpha$ and $w=x^2$ you can reduce to maximize $$|w+2z+2\alpha|^2 + 2|w-z-\alpha|^2 = 6(z\overline\alpha + \overline z\alpha) $$ where we have removed the constant factors coming from $|w|=36$ and $|z|=44$. Adding again $|z|^2$ and $9|\alpha|^2$ you get to maximize $$|z+3\alpha|^2$$ but since $|z|$ is constant, you have the maximum for $z=|z|\alpha/|\alpha|$. Notice that $w$ (and thus $x$) does not affect the problem anymore, so we can fix any $|x|=6$, and then compute $y$ from $z$. retrace back the substitutions to find $u,v$.

Exodd
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