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visits member for 1 year, 5 months
seen Oct 4 '13 at 3:02

Jul
2
awarded  Curious
Apr
30
awarded  Yearling
Oct
4
asked No holomorphic injective function such that $f(B(0,1))=\mathbb{C}$!
Sep
14
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?
Sep
14
comment What is the principal branch of $f(z)=\sqrt{1-z}$, $z\in\mathbb{C}$?
How can it be a log?
Sep
14
asked What is the principal branch of $f(z)=\sqrt{1-z}$, $z\in\mathbb{C}$?
Sep
11
comment Finding holomorphic functions such that $z=(f(z))^n$
In this case it should work because the real part is always positive.
Sep
11
revised Finding holomorphic functions such that $z=(f(z))^n$
edited body
Sep
11
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?
Sep
11
comment Finding holomorphic functions such that $z=(f(z))^n$
And why is that?
Sep
11
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).
Sep
11
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.
Sep
11
revised Finding holomorphic functions such that $z=(f(z))^n$
added 13 characters in body
Sep
11
comment Finding holomorphic functions such that $z=(f(z))^n$
I think it is correct now.
Sep
11
revised Finding holomorphic functions such that $z=(f(z))^n$
added 12 characters in body
Sep
11
comment Finding holomorphic functions such that $z=(f(z))^n$
@JonathanY. could you be more specific?
Sep
11
comment Finding holomorphic functions such that $z=(f(z))^n$
Just fixed it ;)
Sep
11
asked Finding holomorphic functions such that $z=(f(z))^n$
Sep
3
comment Logarithms and the Identity Theorem
Me neither, but thanks anyway :-)
Sep
3
comment Logarithms and the Identity Theorem
But I didn't used the Identity Theorem!