I was wondering: does the function $z^2$ map regions(lets say simply connected regions) ,which are away from the origin, injectively ($1-1$)? Should the region have some extra properties?
Is $z^2$ injective?
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complex-analysis
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1To ensure injectivity the region $D$ must have the property that $z \in D \Rightarrow -z \notin D$. – 2017-01-03
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0Well if $z^2$ maps injectively a region $D$ then that region will have that property, of course, but if it has that property does $z^2$ map it injectively? – 2017-01-03
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1This property is indeed necessary and sufficient. Now, if your region is bounded away from $0$, then convexity is enough. Note that simple-connectedness is not enough, for example you could take $$D= \{ z \in \Bbb{C}: |z|>2 , \mathrm{Im} z - \mathrm{Re} z > 1 \}$$ – 2017-01-03
1 Answers
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$f(z)=z^2$ is injective on $D\subseteq\mathbb{C}\iff(\forall x\neq 0)(x\in D \Rightarrow-x\not\in D)$.
This is immediate from definitions, as $-x$ is the only number that satisfies $P(y):(y\neq x)\land(y^2=x^2)$ in $\mathbb{C}$, and it always (except at zero) satisfies it.
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0If $(z_1)^2=(z_2)^2$ then $(Re(z_1))^2-(Im(z_1))^2=(Re(z_2))^2-(Im(z_2))^2 and Re(z_1)Im(z_1)=Re(z_2)Im(z_2)$.So how can we conclude that $z_1=z_2$?Or that $z_1=-z_2?$ – 2017-01-03
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2$\mathbb{C}$ is a field, so there are at most two solutions to a quadratic equation. Since $a^2=(-a)^2$, those are the only two solutions. – 2017-01-03