For \(0<|z|<\pi/2\), let
\[ f(z)=\frac{z}{\tan z}. \]
Prove that f has a removable singularity at 0.
What is the radius of convergence of the power series for f centered at 0? Justify your answer.
Solution
The extension has value \(f(0)=1\).
1. The quotient extends holomorphically throughout \(|z|<\pi\).
Where the original quotient is defined, \[ f(z)=\frac{z\cos z}{\sin z}. \] The function \(s(z)=\sin z/z\) extends holomorphically to zero with value one, since \[ s(z)=1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots. \] Moreover, the exponential formula for sine gives \(\sin z=0\) exactly when \(e^{2iz}=1\). Writing \(z=x+iy\), the modulus of this equation forces \(y=0\); its argument then gives \(x=k\pi\) for an integer \(k\). Thus \(s\) has no zero in \(|z|<\pi\), including zero. The function \[ F(z)=\frac{\cos z}{s(z)} \] is therefore holomorphic on this whole disk, agrees with \(f\) near zero, and has \(F(0)=1\). This proves (a). By holomorphic implies analytic, the Taylor series of \(F\) at zero converges throughout every closed subdisc of \(|z|<\pi\), so its radius is at least \(\pi\).
2. The Taylor radius at zero is \(\boxed{\pi}\).
At \(z=\pi\), the denominator \(\sin z\) has a simple zero, since \(\cos\pi=-1\ne0\), whereas \(z\cos z=-\pi\ne0\). In particular, \[ \lim_{z\to\pi}(z-\pi)\frac{z\cos z}{\sin z}=\pi\ne0. \] If the Taylor series at zero had radius greater than \(\pi\), its sum would be holomorphic in a neighborhood of \(\pi\) and would agree with \(F\) on \(|z|<\pi\). Continuity would make it bounded as real \(z\uparrow\pi\), contradicting the nonzero limit above. Thus the radius is at most \(\pi\), and hence exactly \(\pi\).
The poles of \(\tan z\) at \(\pm\pi/2\) do not obstruct this Taylor series: the expression \(z\cos z/\sin z\) is holomorphic and zero at those points. They are only missing points of the unsimplified quotient, not poles of the holomorphic germ’s continuation.
3. Q.E.D.
Step 1 proves removability at zero, and step 2 proves that the Taylor radius is \(\pi\).