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Hi guys i want to sketch these set of complex numbers in coordinate system, i hope you can help me.

$a.\{z\in \mathbb{C}||z-1|+|z+1|<4\}$

$b.\{z\in \mathbb{C}| \mathrm{Im}((1-i)z)=0\}$

$c.\{z\in \mathbb{C}|1<|z+3i|<2\}$

Thanks in advance:)

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Note that $|z-a|$ is distance of $z$ from $a$. Use this to translate a and c to geometry questions. –  Maesumi Jan 21 '13 at 3:21
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1 Answer 1

Hint: For (b), $$\text{Im}((1-i)(x+iy))=\text{Im}((x+y)+i(y-x))=(y-x)$$ so what is the area in $\mathbb R^2$ when you are given $$y-x=0$$ For(c): If you set $x+iy=z$ then $|z+3i|=2\equiv|x+i(y+3)|=2~\equiv~\sqrt{x^2+(y+3)^2}=2$ shows a circle in $\mathbb R^2$ centered at $(0,-3)$ with radius $2$. The same is true for $|z+3i|=1$. Now what is $1<|z+3i|<2$? Isn't it the area between two circle which do not contain their borders?

enter image description here

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Nice graphic, again! + 1! –  amWhy Feb 11 '13 at 2:05
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