Exponential form of a complex number

Recall that \(e\) is a mathematical constant approximately equal to 2.71828. You may have seen the exponential function \(e^x = \exp(x)\) for real numbers. It turns out you can extend the domain and the range of the exponential function to the complex exponential function, and we still retain the properties of the original exponential function, namely: $$\frac{d}{dz} e^z = e^z.$$

What happens when the input to the complex exponential function is imaginary?

Euler’s formula. For real \(\theta\), $$e^{i \theta} = \cos{\theta} + i \sin{\theta}.$$

We will informally prove this formula in the problems. Letting \(\theta=\pi\) gives us Euler’s identity:

Euler’s identity. $$e^{i\pi}+1=0.$$

Because of Euler’s formula, we can express any complex number in exponential form \( r e^{i \theta}\), where \(r\) is the modulus and \(\theta\) is an argument. This is essentially equivalent to polar form.


  1. Let \( f(\theta) = \cos{\theta} + i \sin{\theta}\).
    1. Show that \( f'(\theta) / f(\theta) = i\).
    2. By integrating both sides, show that \( f(\theta) = A e^{i \theta} \) for some constant \( A\).
    3. Show that \(f(\theta) = e^{i\theta}\).
  2. Hard. Let \(z = r e^{i\theta}\), where \( -\pi < \theta \le \pi\). We can define \( \text{Log}(z) = \log(r)+ i \theta\).
    1. Evaluate \(\text{Log}(i)\).
    2. Using the change of base formula, show that \(i^i\) is real.