Euler's Formula Evaluator

Math Algebra • Complex Numbers

Written by STEM Calculators Team Published December 20, 2025 Updated February 24, 2026

5. Euler's Formula Evaluator

Evaluate \(e^{i\theta}=\cos\theta+i\sin\theta\), extend to \(e^{a+bi}\), and explore the inverse \( \log(z)\) (principal + branches).

Mode:

Euler evaluation

\(\theta\) input

Accepts pi or π without using eval.

Unit (if no \(\pi\) appears)

If your input contains \(\pi\), it is treated as radians.

Quick preset

Sets \(\theta\) input.

\(\theta\) slider (radians)

Drag for unit-circle motion; it overwrites the input with a numeric radian value.

Options

Show exact values (special angles) Add a constant \((u+vi)\) Animate \(\theta\)

Add constant (so you can do \(e^{i\pi}+1\))

\(u\)

\(v\)

Result shown as \(e^{i\theta} + (u+vi)\).

Unit circle

Complex exponent \(e^{a+bi}\)

\(a\)

\(b\)

Unit for \(b\)

If you want \(b=\pi\), type \(3.14159...\) (or switch to degrees).

Show step-by-step

Inverse (complex logarithm) \(\log(z)\)

Input form

Branch index \(k\)

\(\log_k(z)=\ln r + i(\mathrm{Arg}(z)+2\pi k)\)

Rectangular input

\(x\)

\(y\)

Polar input

\(r\ge 0\)

\(\theta\)

Unit

If \(r=0\), \(\log(z)\) is undefined.

Show step-by-step + branches

Taylor preview (about \(0\))

Show Taylor series preview for \(\sin\theta\) and \(\cos\theta\)

Max power

Tip: see the Taylor Series calculator for more.

Use Fill example for Euler’s identity \(e^{i\pi}+1=0\). For \(\log(z)\), the argument is multi-valued: \(\arg(z)=\mathrm{Arg}(z)+2\pi k,\ k\in\mathbb{Z}\).

Ready

Choose a mode and click Compute.

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Theory: Euler’s Formula, Complex Exponentials, and \(\log(z)\)

1) Euler’s formula

Euler’s formula connects exponentials and trigonometry:

\[ e^{i\theta}=\cos\theta+i\sin\theta. \]

The complex number \(e^{i\theta}\) lies on the unit circle at angle \(\theta\). Its real part is \(\cos\theta\) and its imaginary part is \(\sin\theta\).

Unit circle viewpoint: \(\;|e^{i\theta}|=1\) and \(\arg(e^{i\theta})\equiv \theta\ (\mathrm{mod}\ 2\pi)\).

2) Why it’s true (series idea)

Use Taylor series about \(0\):

\[ e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},\qquad \cos x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!},\qquad \sin x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}. \]

Substitute \(x=i\theta\) into the exponential series and group even/odd powers to obtain \(\cos\theta+i\sin\theta\).

3) Solved example: Euler’s identity

Evaluate \(e^{i\pi}+1\).

\[ e^{i\pi}=\cos\pi+i\sin\pi=-1. \]

\[ e^{i\pi}+1=0. \]

4) Extension: \(e^{a+bi}\)

\[ e^{a+bi}=e^a\,e^{bi}=e^a(\cos b+i\sin b). \]

\[ |e^{a+bi}|=e^a,\qquad \arg(e^{a+bi})\equiv b\ (\mathrm{mod}\ 2\pi). \]

5) Inverse: the complex logarithm \(\log(z)\)

For \(z\neq 0\), write \(z=r\,\mathrm{cis}(\theta)\) with \(r=|z|>0\). Then:

\[ \log(z)=\ln r + i\,\mathrm{Arg}(z). \]

Because \(\theta\) is not unique (\(\theta\to\theta+2\pi k\)), the logarithm is multi-valued:

\[ \log_k(z)=\ln r + i(\theta+2\pi k),\qquad k\in\mathbb{Z}. \]

Frequently Asked Questions

What does Euler’s formula e^(i theta) equal?

Euler’s formula states that e^(i theta) = cos(theta) + i sin(theta). This represents a point on the unit circle at angle theta.

How do you compute e^(a+bi) using Euler’s formula?

Use e^(a+bi) = e^a e^(bi) = e^a(cos b + i sin b). The real part is e^a cos b and the imaginary part is e^a sin b.

How is the complex logarithm log(z) computed and why does it have branches?

For z = r cis(theta) with r > 0, log_k(z) = ln(r) + i(Arg(z) + 2pi k). Because the argument can be shifted by 2pi k, the logarithm is multi-valued and k selects the branch.

When is log(z) undefined in the complex plane?

log(z) is undefined at z = 0 because r = |z| must be positive to take ln(r). If the input has r = 0, there is no valid complex logarithm value.