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Let be given the complex number $z$ satisfying the condition $|z-2+2i|=2\sqrt{2}$. I want to find the complex numbers $z$ so that their modul obtain least and greatest value. I tried. Put $z = x + y i$.

From $|z-2+2i|=2\sqrt{2}$, we have $$|(x-2) + (y+2)i|=2\sqrt{2}.$$ Therefore, $$(x-2)^2 + (y+2)^2=8.$$ And then, I used

Maximize[{x^2 + y^2, (x - 2)^2 + (y + 2)^2 == 8}, {x, y}]

{32, {x -> 4, y -> -4}}

and

Minimize[{x^2 + y^2, (x - 2)^2 + (y + 2)^2 == 8}, {x, y}]

{0, {x -> 0, y -> 0}}

I dont know why I can not use the command

Abs[(x - 2) + (y + 2) I]

to find the modul of (x - 2) + (y + 2) I and how to tell Mathematica to get the result $$(x-2)^2 + (y+2)^2=8$$ when I have $$|(x-2) + (y+2)i|=2\sqrt{2}.$$

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You've missed assumption that x and y are Reals which can be taken into accout by ComplexExpand:

ComplexExpand[Abs[(x - 2) + (y + 2) I]]^2
(-2 + x)^2 + (2 + y)^2

So

Maximize[ComplexExpand /@ {Abs[x + I y], Abs[(x - 2) + (y + 2) I] == 2 Sqrt[2]}, {x, y}]
{4 Sqrt[2], {x -> 4, y -> -4}}
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Does this work for you?

Solve[Abs[z - 2 + 2 I] == 2 Sqrt[2], {z}]
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  • $\begingroup$ No. But I think, the solution does not exactly. $\endgroup$ – minthao_2011 Mar 6 '14 at 4:26

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