$z_1$ and $z_2$ are complex numbers and if $|z_1|=2$ and $(1-i)z_2+(1+i) \bar{z_2}=8 \sqrt{2}$, then prove that minimum value of $|z_1-z_2|=2$

Could someone give me little hint to solve this question?

  • 1
    $\begingroup$ Hint: what geometric curves (in the complex plain) do the two equations prescribe? $\endgroup$ – Fabian Jan 2 '17 at 20:43

Following Fabian's hint, show that

  • $|z_1|=2$ implies $z_1$ is on the circle of radius $2$ centered at the origin, and
  • if you rewrite $(1-i)z_2 + (1+i) \bar{z}-2 = 8\sqrt{2}$ using $z_2=x+iy$, the equation becomes $x+y=4\sqrt{2}$, which is a line in the complex plane with intercepts $4\sqrt{2}$ and $4\sqrt{2}i$.

If you draw a picture of the circle and the line, you can see that the minimizing $z_1$ and $z_2$ lie on the "$45^\circ$" line from the origin, with $|z_1|=2$ and $|z_2|=4$.


You can also go for an algebraic way (if I didn't make a dumb assumption):

Let $z_2=a+bi$

Then $(1-i)(a+bi)+(1+i)(a-bi)=8 \sqrt{2}$ gives us $a+b=4 \sqrt{2} \\\ $

So squaring both sides of $a+b=4 \sqrt{2}$ gives $a^2+b^2+2ab=16(2) \\\ $

Solving this for $a^2+b^2$ gives $a^2+b^2=32-2ab$

So using triangle inequality we can write that: $|z_1-z_2| \ge ||z_1|-|z_2||=|2-\sqrt{32-2ab}|$

Now now we just need to choose $2ab$ so that $|2-\sqrt{32-2ab}|$ will be at it's minimum.


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