The questions asks for me to figure out which point on the line $Re(z)=1$ under the transformation $f(z)=\frac{1}{z}$ was sent to $0$. I know that image is a circle of radius $\frac{1}{2}$ centered at $(\frac{1}{2}, 0)$ however I am confused as to whether $\exists z$ such that $\frac{1}{z} = 0$ as this impossible however since the circle is of radius $\frac{1}{2}$ it leads me to think there must be some point which is mapped there.
Which point on the line $Re(z)=1$ under the transformation $f(z)=\frac{1}{z}$ was sent to $0$.
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complex-analysis
functions
complex-numbers
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0That point would be $\infty$. – 2017-02-11
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0$\lim_{Im(z)\to\infty}$ – 2017-02-11
1 Answers
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Let the point on the line $\operatorname{Re}(z)=1$ be $1+ix$. Then under the transformation $f(z)=\frac 1z$, we observe that $f(1+ix) = \left(\frac{1}{1+x^2}\right)+i\left(\frac{x}{1+x^2}\right) = 0$ suggests $x \rightarrow \infty$