- #1
LagrangeEuler
- 717
- 20
##f(z)=z^{\frac{1}{2}}## has brancing point at ##z=0##.
##z=re^{i\varphi}=re^{i(\varphi+2n\pi)}##
From that
[tex]z^{\frac{1}{2}}=r^{\frac{1}{2}}e^{i (\frac{\varphi}{2}+n\pi)}[/tex]
For ##n=0##, ##z^{\frac{1}{2}}=e^{i \frac{\varphi}{2}}##
For ##n=1##, ##z^{\frac{1}{2}}=e^{i (\frac{\varphi}{2}+\pi)}=-e^{i \frac{\varphi}{2}}##
For ##n=2##, ##z^{\frac{1}{2}}=e^{i \frac{\varphi}{2}}##
For ##n=3##, ##z^{\frac{1}{2}}=e^{i( \frac{\varphi}{2}+\pi)}=-e^{i \frac{\varphi}{2}}##
For ##n=4##, ##z^{\frac{1}{2}}=e^{i \frac{\varphi}{2}}##
##...##
I see that this is multivalued function, but I am not sure why only problem or why only ##z=0## is branching point? Could you please explain me this?
##z=re^{i\varphi}=re^{i(\varphi+2n\pi)}##
From that
[tex]z^{\frac{1}{2}}=r^{\frac{1}{2}}e^{i (\frac{\varphi}{2}+n\pi)}[/tex]
For ##n=0##, ##z^{\frac{1}{2}}=e^{i \frac{\varphi}{2}}##
For ##n=1##, ##z^{\frac{1}{2}}=e^{i (\frac{\varphi}{2}+\pi)}=-e^{i \frac{\varphi}{2}}##
For ##n=2##, ##z^{\frac{1}{2}}=e^{i \frac{\varphi}{2}}##
For ##n=3##, ##z^{\frac{1}{2}}=e^{i( \frac{\varphi}{2}+\pi)}=-e^{i \frac{\varphi}{2}}##
For ##n=4##, ##z^{\frac{1}{2}}=e^{i \frac{\varphi}{2}}##
##...##
I see that this is multivalued function, but I am not sure why only problem or why only ##z=0## is branching point? Could you please explain me this?