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 Jul2 awarded Curious Apr30 awarded Yearling Oct4 asked No holomorphic injective function such that $f(B(0,1))=\mathbb{C}$! Sep14 comment What is the principal branch of $f(z)=\sqrt{1-z}$, $z\in\mathbb{C}$? then it would be $f(z)=\sqrt{|1-z|}e^{i\frac{\theta}{2}}$, where the square root is the positive branch? Sep14 comment What is the principal branch of $f(z)=\sqrt{1-z}$, $z\in\mathbb{C}$? How can it be a log? Sep14 asked What is the principal branch of $f(z)=\sqrt{1-z}$, $z\in\mathbb{C}$? Sep11 comment Finding holomorphic functions such that $z=(f(z))^n$ In this case it should work because the real part is always positive. Sep11 revised Finding holomorphic functions such that $z=(f(z))^n$ edited body Sep11 comment Finding holomorphic functions such that $z=(f(z))^n$ Oohh I think I know. Because I must do $\theta(z)=\arctan\left(\frac{v(z)}{u(z)}\right)+2m\pi$, for some $m\in\mathbb{Z}$, where $z=u(z)+ iv(z)$, and on the non-positive real axis $u$ equals zero? Sep11 comment Finding holomorphic functions such that $z=(f(z))^n$ And why is that? Sep11 comment Finding holomorphic functions such that $z=(f(z))^n$ I still don't figure out why can't z be a negative real number (non-zero). Sep11 comment Finding holomorphic functions such that $z=(f(z))^n$ @JonathanY. I'm so sorry, I forgot to say and edited again; $z$ must belong in $\mathbb{C}-\{z\in\mathbb{R}:Re(z)\leq z\}$, that is, the complex plane minus the non-positive real axis. Sep11 revised Finding holomorphic functions such that $z=(f(z))^n$ added 13 characters in body Sep11 comment Finding holomorphic functions such that $z=(f(z))^n$ I think it is correct now. Sep11 revised Finding holomorphic functions such that $z=(f(z))^n$ added 12 characters in body Sep11 comment Finding holomorphic functions such that $z=(f(z))^n$ @JonathanY. could you be more specific? Sep11 comment Finding holomorphic functions such that $z=(f(z))^n$ Just fixed it ;) Sep11 asked Finding holomorphic functions such that $z=(f(z))^n$ Sep3 comment Logarithms and the Identity Theorem Me neither, but thanks anyway :-) Sep3 comment Logarithms and the Identity Theorem But I didn't used the Identity Theorem!