Find an analytic one-one onto function ${f(z)}$ on $\Bbb{C}$ such that there exist only one $z_{0}$ such that ${f(z_{0})} = z_{0}$.
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7All affine functions are 1-1, and most of them have exactly one fixed point. – Hans Engler Mar 01 '13 at 15:38
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@HansEngler could you elaborate your comment for me? explain and teach me please dear sir – Myshkin Mar 01 '13 at 16:17
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Consider the function $$f(z)=-z$$Then for $z_0=0$ we have $f(z_0)=z_0$ and for all other $z_1\in \mathbb{C}$ we have $f(z_1)=-z_1\neq z_1$. Similarly the function $$f(z)=\pm i z$$ will work. Even more generally any function $$f(z)=e^{i \alpha} z$$ where $0<\alpha < 2 \pi$ will work where in each case $f(z)=z$ iff $z=0$.
Another class of solutions is $$f(z)=\beta z$$ where $\beta \in \mathbb{C}\setminus \{0,1\}$
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