# 2.8: Gallery of Functions

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- 45405

In this section we’ll look at many of the functions you know and love as functions of \(z\). For each one we’ll have to do four things.

- Define how to compute it.
- Specify a branch (if necessary) giving its range.
- Specify a domain (with branch cut if necessary) where it is analytic.
- Compute its derivative.

Most often, we can compute the derivatives of a function using the algebraic rules like the quotient rule. If necessary we can use the Cauchy-Riemann equations or, as a last resort, even the definition of the derivative as a limit.

Before we start on the gallery we define the term “entire function”.

A function that is analytic at every point in the complex plane is called an entire function. We will see that \(e^z\), \(z^n\), \(\sin (z)\) are all entire functions.

## Gallery of functions, derivatives and properties

The following is a concise list of a number of functions and their complex derivatives. None of the derivatives will surprise you. We also give important properties for some of the functions. The proofs for each follow below.

- \(f(z) = e^z = e^x \cos (y) + ie^x \sin (y).\)

Domain = all of \(C\) (\(f\) is entire).

\(f'(z) = e^z\). - \(f(z) \equiv c\) (constant)

Domain = all of \(C\) (\(f\) is entire).

\(f'(z) = 0\). - \(f(z) = z^n\) (\(n\) an integer \(\ge 0\))

Domain = all of \(C\) (\(f\) is entire).

\(f'(z) = nz^{n - 1}\). - \(P(z)\) (polynomial)

A polynomial has the form \(P(z) = a_n z^n + a_{n - 1} z^{n - 1} + ... + a_0\).

Domian = all of \(C\) (\(P(z)\) is entire).

\(P'(z) = na_n z^{n -1} + (n - 1)a_{n -1} z^{n - 1} + ... + 2 a_2 z + a_1.\) - \(f(z) = 1/z\)

Domain = \(C\) - {0} (the punctured plane).

\(f'(z) = -1/z^2.\) - \(f(z) = P(z)/Q(z)\) (rational function)

When \(P\) and \(Q\) are polynomials \(P(z)/Q(z)\) is called a rational function.

If we assume that \(P\) and \(Q\) have no common roots, then:

Domain = \(C\) - {roots of \(Q\)}

\(f'(z) = \dfrac{P'Q - PQ'}{Q^2}.\) - \(\sin (z), \cos (z)\)
Domian: these functions are entire.
\(\cos (z) = \dfrac{e^{iz} + e^{-iz}}{2}\), \(\sin (z) = \dfrac{e^{iz} - e^{-iz}}{2i}\)

(By Euler’s formula we know this is consistent with \(\cos (x)\) and \(\sin (x)\) when \(z = x\) is real.)

\[\dfrac{d \cos (z)}{dz} = -\sin (z), \ \dfrac{d \sin (z)}{dz} = \cos (z).\]

Other key properties of sin and cos:

- \(\cos ^2 (z) + \sin ^2 (z) = 1\)

- \(e^z = \cos (z) + i \sin (z)\)

- Periodic in \(x\) with period \(2\pi\), e.g. \(\sin (x + 2\pi + iy) = \sin (x + iy)\).

- They are not bounded!

- In the form \(f(z) = u(x, y) + iv(x, y)\) we have

\[\cos (z) = \cos (x) \cosh (y) - i \sin (x) \sinh (y) \\ \sin (z) = \sin (x) \cosh (y) + i \cos (x) \sinh (y)\]

(cosh and sinh are defined below.)

- The zeros of \(\sin (z)\) are \(z = n \pi\) for \(n\) any integer.

The zeros of \(\cos (z)\) are \(z = \pi /2 + n\pi\) for \(n\) any integer.

(That is, they have only real zeros that you learned about in your trig. class.) - Other trig functions \(\cot (z)\), \(\sec (z)\) etc.
Domain: The entire plane minus the zeros of the denominator.
The same as for the real versions of these function, e.g. \(\cot (z) = \cos (z) / \sin (z)\), \(\sec (z) = 1/ \cos (z)\).

Derivative: Compute using the quotient rule, e.g.

\[\dfrac{d \tan (z)}{dz} = \dfrac{d}{dz} (\dfrac{\sin (z)}{\cos (z)}) = \dfrac{\cos (z) \cos (z) - \sin (z) (-\sin (z))}{\cos ^2 (z)} = \dfrac{1}{\cos ^2 (z)} = \sec ^2 z\]

(No surprises there!) - \(\sinh (z), \cosh (z)\) (hyperbolic sine and cosine)
Domain: these functions are entire.
\[\cosh (z) = \dfrac{e^z + e^{-z}}{2}, \ \ \sinh (z) = \dfrac{e^z - e^{-z}}{2}\]

\[\dfrac{d \cosh (z)}{dz} = \sinh (z), \ \dfrac{d \sinh (z)}{dz} = \cosh (z)\]

Other key properties of cosh and sinh:

- \(\cosh ^2 (z) - \sinh ^2 (z) = 1\)

- For real \(x\), \(\cosh (x)\) is real and positive, \(\sinh (x)\) is real.

- \(\cosh (iz) = \cos (z)\), \(\sinh (z) = -i \sin (iz).\) - \(\log (z)\) (See Topic 1.)
Branch: Any branch of \(\text{arg} (z)\).
\[\log (z) = \log (|z|) + i \text{arg} (z).\]

Domain: \(C\) minus a branch cut where the chosen branch of \(\text{arg} (z)\) is discontinuous.

\[\dfrac{d}{dz} \log (z) = \dfrac{1}{z}\] - \(z^a\) (any complex \(a\)) (See Topic 1.)
Branch: Any branch of \(\text{arg} (z)\).
\[z^a = e^{a \log (z)}.\]

Domain: Generally the domain is \(C\) minus a branch cut of log. If \(a\) is an integer \(\ge 0\) then \(z^a\) is entire. If \(a\) is a negative integer then \(z^a\) is defined and analytic on \(C\) - {0}.

\[\dfrac{dz^a}{dz} = a z^{a - 1}.\] - \(\sin ^{-1} (z)\)
The definition is chosen so that \(\sin(\sin ^{-1} (z)) = z\). The derivation of the formula is as follows.
\[\sin ^{-1} (z) = -i \log (iz + \sqrt{1 - z^2}).\]

Let \(w = \sin ^{-1} (z)\), so \(z = \sin (w)\). Then,

\[z = \dfrac{e^{iw} - e^{-iw}}{2i} \ \Rightarrow\ e^{2iw} - 2ize^{iw} - 1 = 0\]

Solving the quadratic in \(e^{iw}\) gives

\[e^{iw} = \dfrac{2iz + \sqrt{-4z^2 + 4}}{2} = iz + \sqrt{1 - z^2}.\]

Taking the log gives

\[iw = \log (iz + \sqrt{1 - z^2}) \ \Leftrightarrow \ w = - i \log (iz + \sqrt{1 - z^2}.\]

From the definition we can compute the derivative:

\[\dfrac{d}{dz} \sin ^{-1} (z) = \dfrac{1}{\sqrt{1 - z^2}}.\]

Choosing a branch is tricky because both the square root and the log require choices. We will look at this more carefully in the future.For now, the following discussion and figure are for your amusement.

Sine (likewise cosine) is not a 1-1 function, so if we want \(\sin ^{-1} (z)\) to be single-valued then we have to choose a region where \(\sin (z)\) is 1-1. (This will be a branch of \(\sin ^{-1} (z)\), i.e. a range for the image,) The figure below shows a domain where \(\sin (z)\) is 1-1. The domain consists of the vertical strip \(z = x + iy\) with \(-\pi /2 < x < \pi /2\) together with the two rays on boundary where \(y \ge 0\) (shown as red lines). The figure indicates how the regions making up the domain in the \(z\)-plane are mapped to the quadrants in the \(w\)-plane.

## proofs

Here we prove at least some of the facts stated in the list just above.

- \(f(z) = e^z\). This was done in Example 2.7.1 using the Cauchy-Riemann equations.
- \(f(z) \equiv c\) (constant). This case is trivial.
- \(f(z) = z^n\) (\(n\) an integer \(\ge 0\)): show \(f'(z) = nz^{n - 1}\)

It’s probably easiest to use the definition of derivative directly. Before doing that we note the factorization

\[z^n - z_0^n = (z - z_0) (z^{n - 1} + z^{n - 2} z_0 + z^{n - 3} z_0^2 + ... + z^2 z_0^{n - 3} + zz_0^{n - 2} + z_0^{n - 1})\]

Now

\[\begin{array} {rcl} {f'(z_0)} & = & {\lim_{z \to z_0} \dfrac{f(z) - f(z_0)}{z - z_0} = \lim_{z \to z_0} \dfrac{z^n - z_0^n}{z - z_0}} \\ {} & = & {\lim_{z \to z_0} (z^{n - 1} + z^{n - 2} z_0 + z^{n - 3} z_0^2 + ... + z^2 z_0^{n - 3} + zz_0^{n - 2} + z_0^{n - 1}} \\ {} & = & {nz_0^{n - 1}.} \end{array}\]

Since we showed directly that the derivative exists for all \(z\), the function must be entire. - \(P(z)\) (polynomial). Since a polynomial is a sum of monomials, the formula for the derivative follows from the derivative rule for sums and the case \(f(z) = z^n\). Likewise the fact the \(P(z)\) is entire.
- \(f(z) = 1/z\). This follows from the quotient rule.
- \(f(z) = P(z)/Q(z)\). This also follows from the quotient rule.
- \(\sin (z)\), \(\cos (z)\). All the facts about \(\sin (z)\) and \(\cos (z)\) follow from their definition in terms of exponentials.
- Other trig functions \(\cot (z)\), \(\sec (z)\) etc. Since these are all defined in terms of cos and sin, all the facts about these functions follow from the derivative rules.
- \(\sinh (z)\), \(\cosh (z)\). All the facts about \(\sinh (z)\) and \(\cosh (z)\) follow from their definition in terms of exponentials.
- \(\log (z)\). The derivative of \(\log (z)\) can be found by differentiating the relation \(e^{\log (z)} = z\) using the chain rule. Let \(w = \log (z)\), so \(e^w = z\) and

\[\dfrac{d}{dz} e^w = \dfrac{dz}{dz} = 1 \ \Rightarrow \ \dfrac{de^w}{dw} \dfrac{dw}{dz} = 1 \ \Rightarrow \ e^w \dfrac{dw}{dz} = 1 \ \Rightarrow \ \dfrac{dw}{dz} = \dfrac{1}{e^w}\]

Using \(w = \log (z)\) we get

\[\dfrac{d \log (z)}{dz} = \dfrac{1}{z}.\] - \(z^a\) (any complex \(a\)). The derivative for this follows from the formula

\[z^a = e^{a \log (z)} \ \Rightarrow \ \dfrac{dz^a}{dz} = e^{a \log (z)} \cdot \dfrac{a}{z} = \dfrac{az^a}{z} = az^{a - 1}\]