- #1
latentcorpse
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i need to check [itex]z^a[/itex] is holomorphic on [itex]\mathbb{C} \backslash \{0\}, a \in \mathbb{C}[/itex] but am having difficulty arranging it into a form that i can use the Cauchy Riemann equations on. so far is have:
[itex]z^a=e^{a \ln{z}}=e^{a \ln{|z|}}e^{i arg(z)}[/itex]
which when i break up the second exponential gives
[itex]z^a=e^{a \ln{(x^2+y^2)^{\frac{1}{2}}}} \cos{\arctan{(\frac{y}{x})}}+i e^{a \ln{(x^2+y^2)^{\frac{1}{2}}}} \sin{\arctan{(\frac{y}{x})}}[/itex]
so if i let [itex]z=u(x,y)+i v(x,y)[/itex] we get
[itex]u(x,y)=e^{a \ln{(x^2+y^2)^{\frac{1}{2}}}} \cos{\arctan{(\frac{y}{x})}}[/itex]
[itex]v(x,y)=e^{a \ln{(x^2+y^2)^{\frac{1}{2}}}} \sin{\arctan{(\frac{y}{x})}}[/itex]
[itex]z^a=e^{a \ln{z}}=e^{a \ln{|z|}}e^{i arg(z)}[/itex]
which when i break up the second exponential gives
[itex]z^a=e^{a \ln{(x^2+y^2)^{\frac{1}{2}}}} \cos{\arctan{(\frac{y}{x})}}+i e^{a \ln{(x^2+y^2)^{\frac{1}{2}}}} \sin{\arctan{(\frac{y}{x})}}[/itex]
so if i let [itex]z=u(x,y)+i v(x,y)[/itex] we get
[itex]u(x,y)=e^{a \ln{(x^2+y^2)^{\frac{1}{2}}}} \cos{\arctan{(\frac{y}{x})}}[/itex]
[itex]v(x,y)=e^{a \ln{(x^2+y^2)^{\frac{1}{2}}}} \sin{\arctan{(\frac{y}{x})}}[/itex]