Home » Blog » Prove some properties of complex numbers raised to complex powers

Prove some properties of complex numbers raised to complex powers

If z is a nonzero complex numbers and w \in \mathbb{C} let

    \[ z^w = e^{w \operatorname{Log} z}, \]

where

    \[ \operatorname{Log} z = \log |z| + i \arg(z). \]

  1. Compute 1^i, i^i, and (-1)^i.
  2. Prove that z^a z^b = z^{a+b} if a,b, and z are in \mathbb{C} with z \neq 0.
  3. What conditions on z_1 and z_2 must we have for the equation

        \[ (z_1 z_2)^w = z_1^w z_2^w \]

    to hold? Show that the equation fails when z_1 = z_2 = -1 and w = i.


  1. The computations are as follows,

        \begin{align*}  1^i &= e^{i \operatorname{Log} 1} = e^0 = 1. \\[9pt]  i^i &= e^{i \operatorname{Log} i} = e^{i \left( \frac{\pi i}{2} \right)} = e^{-\frac{\pi}{2}}. \\[9pt]  (-1)^i &= e^{i \operatorname{Log} (-1)} = e^{i (\pi i)} = e^{- \pi}.  \end{align*}

  2. Proof. Using the definitions we compute,

        \begin{align*}  z^a z^b &= \left( e^{a \operatorname{Log} z} \right) \left( e^{b \operatorname{Log} z} \right) \\[9pt]  &= e^{a \operatorname{Log} z + b \operatorname{Log} z} \\[9pt]  &= e^{(a+b)\operatorname{Log} z} \\[9pt]  &= z^{a+b}. \qquad \blacksquare \end{align*}

  3. First, if z_1 = z_2 = -1 and w = i then we have

        \[ (z_1 z_2)^w = 1^i = 1, \]

    but

        \[ z_1^w z_2^w = (-1)^i \cdot (-1)^i = e^{-\pi} \cdot e^{-\pi} = e^{-2\pi}. \]

    Thus,

        \[ (z_1 z_2)^w \neq z_1^w z_2^w. \]

    In order for (z_1 z_2)^w = z_1^w z_2^w we must have - \pi < \arg z_1 + \arg z_2 \leq \pi since

        \[ (z_1 z_2)^w = e^{w \operatorname{Log} (z_1 z_2)} = z_1^w z_2^w \cdot e^{2n \pi i}, \]

    and n = 0 only when - \pi < arg z_1 + arg z_2 \leq \pi.

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):