
### Tutorial 10 (Week 11)

Not all examples should be covered.

### 3.1: Theorems

THEOREM 1. Suppose that $h$ is analytic in a domain $D$ except for a finite number of poles. Let $\gamma$ be a piecewise smooth positively oriented simple closed curve in $D$, which does not pass through any pole or zero of h and whose inside lies in $D$. Then, number of zeros) fnumber of poles $$\frac{1}{ 2\pi i}\int_\gamma \frac{h'(z)}{h(z)}\,dz =N(h,D)-P(h,D) \label{eqn-1}$$ where $N(h,D)$ is the number of zeros of $h$ inside $D$ and $P(h,D)$ is the number of poles of $h$ inside $D$. All zeros and poles are counted with multiplicities.

THEOREM 1'. The same is true for negatively oriented curve $\gamma$ and external domain $D$.

THEOREM 2 (The Argument Principle). In the framework of Theorem 1 or Theorem 2 expression (\ref{eqn-1}) equals to the change in $\arg(h(z))$ as $z$ traverses $\gamma$, multiplied by $\frac{1}{2\pi}$.

LEMMA. Theorem Suppose $f$ and $g$ are contionuous on a piecewise smooth simple closed curve $\gamma$. Let $$|f(z)-g(z)|<|f(z)|\qquad \forall z\in \gamma. \tag{5}\label{eqn-5}$$ Then change in $\arg(f(z))$ as $z$ traverses $\gamma$ equals to change in $\arg(g(z))$ as $z$ traverses $\gamma$.,

THEOREM 3 (Rouche's). Theorem Suppose $f$ and $g$ are analytic on an open set containing a piecewise smooth simple closed curve $\gamma$ and its inside. Let (\ref{eqn-5}) be fulfilled. Then $f$ and $g$ have an equal number of zeros inside $\gamma$, counting multiplicities.

THEOREM 3'. Thesame is true for outside domain.

THEOREM 4 (The Fundamental Theorem of Algebra). A polynomial of degree $n$ has exactly $n$ zeros, counting multiplicities.

### Problems

Use the technique of Example 2 to determine the number of zeros of $f(z) = z^2 - z + 1$ in the first quadrant.

### 3.2: Theorems

THEOREM 1. Suppose that $f$ is a nonconstant analytic function on a domain $D$. Then the range of $f(z)$, as $z$ varies over $D$, is an open set.

COROLLARY 1 (The Maximum-Modulus Principle). If $f$ is a nonconstant analytic function on a domain $D$, then $|f|$ can have no local maximum on $D$.

THEOREM 2 (Schwarz's Lemma). Suppose that $f$ is analytic in the disc ${|z| < 1}$, that $f(0) = 0$, and that $|f(z)| \le 1$, for all $z$ in the disc. Then $|f(z)| \le |z|$ for all $z$ in the disc.

### Problems

1. Let $f(z) = z^2/(z+2)$; find the maximum value of $|f(z)|$ as $z$ varies over the disc ${|z|\le 1}$.

2. Let $f(z) = ze^z$; find the maximum value of |f(z)| as $z$ varies over the region $D = {x + iy: x^2 + y^2 \le 4, x \ge 0, y \ge 0}$.