Transcendental Equations in a Complex Variable

Level 1 - Math I (Physics) topic page in Complex Equations.

Principle

A transcendental equation is an equation involving functions that are not finite polynomials or rational functions of the unknown. Complex exponentials and trigonometric functions produce many such equations.

Transcendental equation

\[f(z)=0\quad\text{with }f\text{ not a polynomial or rational function}\]

The fundamental theorem of algebra does not control transcendental equations. They may have no solutions, finitely many solutions, or infinitely many solutions.

Notation

\(z=x+iy\) is the unknown complex number.
\(x=\operatorname{Re}(z)\) and \(y=\operatorname{Im}(z)\).
\(k\) is an integer used to record periodic solutions.
\(e^z=e^x e^{iy}\) is the complex exponential.

Method

When an equation contains \(e^z\), write \(z=x+iy\):

Exponential split

\[e^z=e^xe^{iy}=e^x(\cos y+i\sin y)\]

Then match modulus and argument. For example, solving \(e^z=1\) gives:

Basic equation

\[e^xe^{iy}=1=e^{2\pi i k}\]

Matching moduli gives \(e^x=1\), so \(x=0\). Matching arguments gives \(y=2\pi k\). Therefore:

Basic solution

\[z=2\pi i k,\quad k\in\mathbb Z\]

Rules

Always include the periodic term \(2\pi k\) when equating complex arguments.
The equation \(e^z=0\) has no solution because \(|e^z|=e^x\gt0\).
Trigonometric and hyperbolic equations can often be rewritten as exponential equations.
Solution sets can be infinite because complex exponentials are periodic in the imaginary direction.

Examples

Solve \(\sinh z=0\):

Sinh equation

\[\sinh z=\frac{e^z-e^{-z}}{2}=0\]

Multiply by \(2e^z\):

Sinh exponential equation

\[e^{2z}=1\]

Use the solution of \(e^w=1\):

Sinh solution

\[2z=2\pi i k\quad\Longrightarrow\quad z=\pi i k,\quad k\in\mathbb Z\]

Solve \(\cos z=0\):

Cos equation

\[\cos z=\frac{e^{iz}+e^{-iz}}{2}=0\]

Multiply by \(2e^{iz}\):

Cos exponential equation

\[e^{2iz}=-1=e^{i(\pi+2\pi k)}\]

Therefore:

Cos solution

\[2iz=i(\pi+2\pi k)\quad\Longrightarrow\quad z=\pi\left(k+\frac12\right),\quad k\in\mathbb Z\]

Solve \(e^z=1+i\):

Target polar form

\[1+i=\sqrt2e^{i(\pi/4+2\pi k)}\]

Matching modulus and argument gives:

Exponential target solution

\[z=\frac12\ln2+i\left(\frac\pi4+2\pi k\right),\quad k\in\mathbb Z\]

Finally, \(e^z=0\) has no solution because matching moduli would require \(e^x=0\), which never happens for real \(x\).

Checks

Do not apply the fundamental theorem of algebra to transcendental equations.
Include all integer values of \(k\) when periodicity creates infinitely many solutions.
Check whether the equation has no solution by taking moduli.
When rewriting trigonometric functions as exponentials, multiply by a non-zero exponential factor such as \(e^z\) or \(e^{iz}\).

Complex Functions

Symmetries