Functions of the Complex Variable

Analytic Polynomials

Definition: A polynomial $P(x, y)$ is an analytic polynomial if there exist complex constants $a_k$ such that

\[P(x, y) = a_0 + a_1(x + iy) + a_2(x + iy)^2 + \dots + a_N(x + iy)^N\]

Then $P$ is polynomial in $z$ and write it as

\[P(z) = a_0 + a_1z + \dots + a_Nz^N\]

Example: $x^2 + iv(x, y)$ is not analytic for any choice of the real polynomial $v(x, y)$, since a polynomial in $z$ can have a real part of degree 2 in $x$ if and only if it is of the form $az^2 + bc + c$ with $a \neq 0$, in which case the real part must contain a $y^2$ term as well.

Definition: Let $f(x, y) = u(x, y) + iv(x, y)$ where $u$ and $v$ are real-valued functions. The partial derivatives $f_x$ and $f_y$ are defined by $u_x + iv_x$ and $u_y + iv_y$ respectively, provided the latter exist.

Proposition: A polynomial $P(x, y)$ is analytic if and only if $P_y = iP_x$.

Proof: $(\rightarrow)$ Suppose $P = \sum_{k = 1}^N a_k (x + iy)^k$ is analytic. Then $P_y = i\sum_{k = 1}^N a_kk(x + iy)^{k - 1}$ and $P_x = \sum_{k = 1}^N a_kk(x + iy)^{k - 1}$. So $P_y = iPx$.

$(\leftarrow)$ TODO. (Sketch: Compute $P_y$ and $P_x$ and compare coefficients to show that $C_k = i^k\binom{n}{k}C_0$, so that $P(x, y) = \sum_{k = 0}^n C_kx^{n - k}y^k = C_0\sum_{k = 0}^n \binom{n}{k}x^{n - k}(yi)^k = C_0(x + iy)^n$). $\blacksquare$

Remark: Since $f_x = u_x + iv_x$ and $f_y = u_y + iv_y$, $f_y = if_x$ is equivalently

\[\begin{align*} u_x &= v_y \\ -v_x &= u_y \end{align*}\]

these are the Cauchy-Riemann equations.

Example: By the Cauchy-Riemann equations, a non-constant analytic polynomial cannot be real-valued, since if it were both $f_x$ and $f_y$ would be real, so $f_y = if_x$ could not hold.

Definition: A complex-valued function, defined in a neighborhood of $z$, is said to be differentiable if

\[\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\]

exists. The value of the limit is denoted $f’(z)$.

Proposition: The sum, difference, product, and quotient of functions differentiable at $z$ is differentiable at $z$ (assuming the denominator at $z$ is non-zero).

Proposition: The power rule holds for complex polynomials as well.

Power Series

Definition: A power series is an infinite series in the form $\sum_{k = 1}^N C_kz^k$.

Definition: The limit supremum of a real-valued sequence $(a_n)$ is defined as

\[\limsup_{n \to \infty} a_n = \lim_{n \to \infty} \left( \sup_{k \geq n} a_k \right)\]

Remark: Note that $\limsup$ always exists or equals $\infty$ since $\sup_{n \geq k} a_k$ is a non-increasing function of $n$.

Proposition: Let $\limsup_{n \to \infty} a_k = L$

1. For all $N$ and for all $\epsilon$, there exists a $k \gt N$ such that $a_k \geq L - \epsilon$ (there are infinitely many $a_k$ that are $\epsilon$-close to the limit)
2. For all $\epsilon$, there exists some $N$ such that for all $k \gt N$, $a_k \leq L + \epsilon$ (the sequence gets arbitrary close to the limit)
3. $\limsup ca_n = cL$ for any non-negative constant $c$.

Theorem: Suppose $\limsup \vert C_k \vert^{1/k} = L$. Let $S = \sum C_k z^k$. Then

1. If $L = 0$, $S$ converges for all $z$.
2. If $L = \infty$, $S$ converges for $z = 0$ only.
3. If $0 \lt L \lt \infty$, set $R = 1/L$. $S$ converges when $\vert z \vert \lt R$ and diverges when $\vert z \vert \gt R$. $R$ is called the radius of convergence.

Proof:

1. Suppose $L = 0$. Then $\limsup \vert C_k \vert^{1/k} \vert z \vert = 0$ for all $z$. So, $\vert C_k z^k \vert \leq \frac{1}{2^k}$ for all $k$ greater than some $N$. So, $\sum \vert C_k z^k \vert$ converges by the Comparison Test, so $S$ converges by the Absolute Convergence Test.
2. Suppose $L = \infty$. Then $\vert C_k \vert^{1/k} \geq \frac{1}{\vert z \vert}$ for all $z \neq 0$. So $\vert C_k z^k \vert \geq 1$, so the terms of the series do not approach 0, so $S$ diverges. If $z = 0$, then every term of the series is 0 so $S$ converges.
3. First suppose $\vert z \vert \lt R$. Choose $\delta$ such that $\vert z \vert = R(1 - 2\delta)$. Then $\limsup \vert C \vert^{1/k} \vert z \vert = 1 - 2\delta$, so $\vert C \vert\vert z \vert^k < (1 - \delta)^k$. So $S$ is absolutely convergent, so $\sum C_k z^k$ is convergent. Now suppose $\vert z \vert \gt R$. Then $\limsup \vert C_k \vert^{1/k} \vert z \vert \gt 1$, so $\vert C_kz^k \vert \gt 1$, so the terms of $S$ do not converge to 0, so $S$ diverges. $\blacksquare$

Remark: The theorem says nothing about when $\vert z \vert = R$; indeed, when this is the case the series can converge or diverge.

Proposition: The sum of two power series is convergent when each of the two power series are themselves convergent.

Differentiability and Uniqueness of Power Series

Proposition: Let $\sum C_n z^n$ be convergent in a disc $D(0; R), R \gt 0$. Then the series $\sum n C_n z^{n - 1}$ obtained by differentiating the first series term by term is also convergent in a disc $D(0; R)$.

Proof: $\limsup \vert nC_n \vert^\frac{1}{n - 1} = \limsup (\vert nC_n \vert^\frac{1}{n})^\frac{n}{n - 1} = \limsup \vert C_n \vert^\frac{1}{n}$ $\blacksquare$

Theorem: Suppose $f(z) = \sum C_n z^n$ converges for $\vert z \vert \lt R$. Then $f’(z)$ exists and equals $\sum nC_n z^{n - 1}$ exists when $\vert z \vert \lt R$.

Proof: TODO.

Corollary: Power series are infinitely differentiable within their domain of convergence.

Proof: TODO. (Sketch: Using induction: use the above theorem to show that $f(z)$ is differentiable once, and since the radius of convergence is the same by the above proposition, it can be differentiated again.)

Corollary: If $f(z) = \sum C_nz^n$ has a non-zero radius of convergence,

\[C_n = \frac{f^{(n)}(0)}{n!}\]

Proof: $f^{(n)}(z) = n!C_nz^0 + n!C_{n + 1}z + \dots$, so $f^{(n)}(0) = n!C_nz^0$, from which the corollary follows. $\blacksquare$

Theorem (Uniqueness Theorem for Power Series): Suppose $\sum C_n z^n$ is zero at all points of a non-zero sequence $(z_k)$ convergent to 0. Then the series is identically 0.

Proof: Let $f(z) = \sum C_nz^n$. Then by continuity of $f$ at the origin,

\[C_0 = f(0) = \lim_{z \to 0} f(z) = \lim_{k \to \infty}(z_k) = 0\]

and similarily for $g(z) = f(z)/z$, $C_1 = 0$, et cetera. So all the coefficients are 0, so the power series is identically 0. $\blacksquare$

Corollary: If a power series equals 0 at all points of a set with an accumulation point at the origin, then the power series is identically 0.

Proof: Since there is an accumulation point at the origin, we can define a sequence $(z_n)$ convergent to 0, from which the above theorem follows. $\blacksquare$

Corollary: If $\sum a_n z^n$ and $\sum b_n z^n$ agree on a set of points with an accumulation point at the origin, then $a_n = b_n$ for all $n$.

Proof: Consider the series $\sum (a_n - b_n) z_n$, from which the above corollary can be applied to show that $a_n - b_n = 0$ for all $n$. $\blacksquare$