Complex functions and polynomials: Complex functions
The complex exponential function
So far, we have studied how to compute powers of complex numbers, like #\blue{z}^n# for real #n#. We will now switch roles and learn what numbers of the form #t^\blue{z}# look like for real #t# and complex #\blue{z}#. We will first focus on the #t=\e# case, which will be our first example of an exponential function of complex numbers.
The complex exponential function
Let #\blue{z}=\green{a}+\orange{b}\cdot\complexi#. The complex exponential function is defined as
It follows that the norm, #\norm{\e^{\blue{z}}}#, and angle, #\theta#, of #\e^{\blue{z}}# are given by #\e^{\green{a}}# and #\orange{b}#, respectively. \[\begin{array}{rcl}\e^{\blue{z}}&=&\e^{\green{a}}\cdot\left(\cos(\orange{b})+\sin(\orange{b})\cdot\complexi\right)\\ &=&\norm{\e^{\blue{z}}}\cdot\left(\cos\left(\theta\right)+\sin\left(\theta\right)\cdot\complexi\right)\end{array}\]
where #\theta# denotes the angle of the number #\e^{\blue{z}}#, not the angle of #\blue{z}#.
This video gives an example of how to calculate the exponential of a complex number #\blue z#, and states some calculation rules for complex exponents.
The voice in the video is AI-generated and not a human voice.
In the visualisations, we can see what happens to #\e^{\blue{z}}# when we fix the value of #\green a# or #\orange b# and vary the other.
Examples
Example 1
Consider #\blue{z}=\orange{\pi}\cdot\complexi#. Then, #\green{a}=\green{0}# and #\orange{b}=\orange{\pi}#. It follows that \[\begin{array}{rcl} \e^{\blue{z}} \; = \; \e^{\orange{\pi}\cdot\complexi} &=&\e^{\green{0}} \cdot \left(\cos\left(\orange{\pi}\right)+\sin\left(\orange{\pi}\right)\cdot\complexi\right)\\ &=&-1\end{array}\]
This is known as Euler's identity, often written as #\;\e^{\orange{\pi}\cdot\complexi}+1=0#.
Example 2
Consider #\e^{\blue{z}}=1+\complexi#. We can write the right-hand side in polar form, obtaining \[ 1+\complexi = \green{\sqrt{2}}\cdot\left(\cos\left(\orange{\frac{\pi}{4}}\right)+\sin\left(\orange{\frac{\pi}{4}} \right)\cdot\complexi\right)\]
and #\blue{z}# is given by \[\blue{z}=\green{\ln\left(\sqrt{2}\right)}+\orange{\frac{\pi}{4}}\cdot \complexi\]
Visualisation complex exponentiation I
In this visualiser, we have three points with equal #\green{a}# and different #\orange{b}#. Since they only differ in the #\orange{b}# value, their exponentials have the same size and lie on a circle of radius #\e^{\green{a}}#.
In this visualiser, we have three points with equal #\orange{b}# and different #\green{a}#. Since they only differ in the #\green{a}# value, their exponentials all have the same argument and different magnitudes.
Note that if #\orange{\mathrm{Im}(z)}=\orange{0}#, then #\mathrm{Arg}\left(\e^{\blue{z}}\right)=0# and #\e^{\blue{z}}# lies on the positive real axis, coinciding with the exponential of the real number #\blue{z}#.
If #\blue{z}# is purely imaginary, then #\blue{z}=\orange{b}\cdot\complexi# and thus #\norm{\e^{\blue{z}}}=1#, meaning #\e^{\blue{z}}# is on the unit circle on the complex plane with the value of #\orange{b}# determining the angle with the real positive axis.
We recognise the polar form in the definition of the complex exponential function. We now consider some computational rules for complex exponentials.
Rules for computing with complex exponents
The exponential of a complex number satisfies the following properties:
For this proof, we will use the fourth rule and the rule that any (complex) number divided by itself equals #1#. We start by writing #\blue 0 =z-z#. \[\e^\blue{0}=\e^{z-z} = \dfrac{\e^z}{\e^z}=1 \]which concludes the proof.
Proof of #\e^{-\blue z} = \dfrac{1}{\e^\blue z}#
Let #\blue{z}=\green{a} +\orange{ b}\cdot\complexi#, then #-\blue{z}=-\green{a} -\orange{ b}\cdot\complexi#.
First, we rewrite the left-hand side.\[\begin{array}{rcl} \e^{-\blue{z}} &=&\e^{-\green{a}} \cdot \left(\cos\left(-\orange{b}\right)+\sin\left(-\orange{b}\right)\cdot \ii \right) \\ &&\quad\blue{\text{used the definition of the complex exponential}}\\ &=&\dfrac{1}{\e^{\green a}} \cdot \left(\cos\left(\orange{b}\right)-\sin\left(\orange{b}\right)\cdot \ii\right) \\ &&\quad \blue{\text{used } x^{-n}=\frac{1}{x^n} \text{ with } x \text{ and } n \text{ real, and } }\\ &&\quad \blue{\cos(- \alpha)=\cos(\alpha)\text{ and }\sin(-\alpha)=-\sin(\alpha)} \end{array} \]
Then, we rewrite the right-hand side.\[\begin{array}{rcl} \dfrac{1}{\e^{\blue z}} &=&\dfrac{1}{\e^{\green{a}} \cdot \left(\cos\left(\orange{b}\right)+\sin\left(\orange{b}\right)\cdot \ii\right)} \\ &&\quad\blue{\text{used the definition of the complex exponential}}\\ &=&\dfrac{\cos\left(\orange{b}\right)-\sin\left(\orange{b}\right)\cdot \ii}{\e^{\green a} \cdot \left( \cos\left(\orange{b}\right)+\sin\left(\orange{b}\right)\cdot \ii \right) \cdot \left( \cos\left(\orange{b}\right)-\sin\left(\orange{b}\right)\cdot \ii \right)}\\ &&\quad \blue{\text{multiplied and divided by the conjugate of }\cos(b)+\sin(b) \cdot \ii}\\ &=&\dfrac{\cos\left(\orange{b}\right)-\sin\left(\orange{b}\right)\cdot \ii}{ \e^{\green a} \cdot \left( \left( \cos(\orange{b}) \right)^2 + \left( \sin(\orange{b}) \right)^2 \right)}\\ &&\quad \blue{\text{expanded the brackets}}\\ &=&\dfrac{1}{\e^{\green a}} \cdot \left(\cos\left(\orange{b}\right)-\sin\left(\orange{b}\right)\cdot \ii\right) \\ &&\quad \blue{\text{simplified using } \left( \sin(\alpha) \right)^2 + \left( \cos(\alpha) \right)^2 = 1}\\ \end{array}\]
Finally, we see that both left-hand and right-hand sides can be written in the same way, which proves the equation.
Proof of #\e^{\blue{z}}\cdot \e^{\purple{w}}=\e^{\blue{z}+\purple{w}}#.
Let #\blue{z}=\green{a}+\orange{b}\cdot\complexi# and #\purple{w}=\green{c}+\orange{d}\cdot\complexi#. We know that #\blue{z}+\purple{w}=\left(\green{a}+\green{c}\right)+\left(\orange{b}+\orange{d}\right)\cdot\complexi#. Therefore, using the definition of complex exponential \[\e^{\blue{z}+\purple{w}}=\e^{\green{a}+\green{c}}\cdot\left(\cos\left(\orange{b}+\orange{d}\right)+\sin\left(\orange{b}+\orange{d}\right)\cdot\complexi\right)\] We now proceed to show that #\e^{\blue{z}}\cdot \e^{\purple{w}}# equals the right-hand side of the previous equation by using the definition of complex exponential and trigonometric identities \[\begin{array}{rcl} \e^{\blue{z}}\cdot \e^{\blue{w}}&=& \left(\e^{\green{a}+\orange{b}\cdot\complexi}\right) \cdot\left(\e^{\green{c}+\orange{d}\cdot\complexi}\right)\\ &&\quad\blue{\text{wrote the complex numbers }z\text{ and }w\text{ in terms of their real and imaginary parts}}\\ &=& \left(\e^{\green{a}}\cdot\left(\cos\left(\orange{b}\right)+\sin\left(\orange{b}\right)\cdot\complexi\right)\right) \cdot \left(\e^{\green{c}}\cdot\left(\cos\left(\orange{d}\right)+ \sin\left(\orange{d}\right)\cdot\complexi\right)\right)\\ &&\quad\blue{\text{used the definition of the complex exponential}}\\ &=& \e^{\green{a}+\green{c}}\cdot\left(\cos\left(\orange{b}\right)\cdot\cos\left(\orange{d}\right)-\sin\left(\orange{b}\right)\cdot\sin\left(\orange{d}\right)+ \left(\cos\left(\orange{b}\right)\cdot\sin\left(\orange{d}\right)+\sin\left(\orange{b}\right)\cdot\cos\left(\orange{d}\right)\right)\cdot\complexi\right)\\ &&\quad\blue{\text{used }\e^x\cdot\e^y=\e^{x+y}\text{ for real numbers and expanded the trigonometric terms}}\\ &=&\e^{\green{a}+\green{c}}\cdot\left(\cos\left(\orange{b}+\orange{d}\right)+\sin\left(\orange{b}+\orange{d}\right)\cdot\complexi\right)\\ &&\quad\blue{\text{used }\cos(\alpha+\beta)=\cos(\alpha)\cdot\cos(\beta)-\sin(\alpha)\cdot\sin(\beta)}\\ &&\quad\blue{\text{ and }\sin(\alpha+\beta)=\cos(\alpha)\cdot\sin(\beta)+\cos(\beta)\cdot\sin(\alpha)}\end{array}\]which is exactly what we wanted to prove.
Proof of #\frac{\e^{\blue{z}}}{\e^{\purple{w}}}=\e^{\blue{z}-\purple{w}}#.
This automatically follows from the second and third rules. \[\dfrac{\e^{\blue{z}}}{\e^{\purple{w}}}=\e^{\blue{z}} \cdot \dfrac{1}{\e^{\purple{w}}}= \e^{\blue z} \cdot \e^{- \purple w} =\e^{\blue z - \purple w} \]
Proofs rule 5 - 6
Proof of #\left(\e^{\blue{z}}\right)^n=\e^{\left(\blue{z}\cdot n\right)}#.
We will proceed by induction. The first step is to show that the equation holds for the base case, #n=1#. In that case, we have #\e^{\blue{z}}=\e^{\blue{z}}#, which is clearly true. Next, we assume the equation is true for #n=k#, the induction hypothesis. Now, we write the equation for #n=k+1#, and using the induction hypothesis, we show it is true, and thus show that it is true for all #n#, \[\begin{array}{rcl} \displaystyle \left(\e^{\blue{z}}\right)^{k+1}&=& \displaystyle\e^{\blue{z}\cdot\left(k+1\right)}\\ &&\quad\blue{\text{wrote the equation for }n=k+1}\\ \displaystyle\left(\e^{\blue{z}}\right)^k\cdot\left(e^{\blue{z}}\right)&=&\displaystyle\e^{\blue{z}\cdot k+\blue z}\\ &&\quad\blue{\text{used }y^{a+b}=y^a\cdot y^b\text{ on the left-hand side for }y=\e^z}\\ &&\quad\blue{\text{and expanded the exponent of the right-hand side}}\\ \displaystyle\left(\e^{\blue{z}}\right)^k\cdot\left(\e^{\blue{z}}\right)&=&\displaystyle\e^{\blue{z}\cdot k}\cdot\e^{\blue{z}}\\ &&\quad\blue{\text{used }\e^{a+b}=\e^a\cdot\e^b\text{ on the right-hand side}}\\ \left(\e^{\blue{z}}\right)^k\cdot \e^{\blue{z}} &=& \left(\e^{\blue{z}}\right)^k\cdot \e^{\blue{z}}\\ &&\quad\blue{\text{used the induction hypothesis, i.e. }\left(\e^z\right)^k=\e^{z\cdot k} \text{ on the right-hand side}}\end{array}\]
which finishes the proof.
Proof of #\overline{\,\e^{\blue{z}}}=\e^{\blue{\overline{z}}}#.
Let #\blue{z}=\green{a} +\orange{ b}\cdot\complexi#. Recall #\blue{\overline{z}}=\green{a}-\orange{b}\cdot\complexi# and #\e^{\blue{z}}=\e^{\green{a}}\cdot\left(\cos(\orange{b})+\sin(\orange{b})\cdot\complexi\right)#. Let us now compute the expressions on both sides of the equation separately. \[\begin{array}{rcl}\overline{\,\e^{\blue{z}}} &=& \overline{\e^{\green{a}}\cdot\left(\cos\left(\orange{b}\right)+\sin\left(\orange{b}\right)\cdot\complexi\right)}\\ &&\quad\blue{\text{used the definition of the complex exponential function }\e^z}\\ &=&\e^{\green{a}}\cdot\left(\cos\left(\orange{b}\right)-\sin\left(\orange{b}\right)\cdot\complexi\right) \\&&\quad\blue{\text{used the definition of complex conjugation, switched the sign of the imaginary part}}\end{array}\]
and \[\begin{array}{rcl}\e^{\blue{\overline{z}}}&=&\e^{\green{a}}\cdot \left(\cos\left(\orange{-b}\right)+\sin\left(\orange{-b}\right)\cdot\complexi \right)\\ &&\quad\blue{\text{applied the definition of complex exponentiation to the number }\bar{z}}\\ &=&\e^{\green{a}}\cdot \left(\cos(\orange{b})-\sin(\orange{b})\cdot\complexi \right)\\ &&\quad\blue{\text{used }\cos(- \alpha)=\cos(\alpha)\text{ and }\sin(-\alpha)=-\sin(\alpha)} \end{array}\] The last step ensured that the right-hand side of #\e^{\blue{\overline{z}}}# is the same as the right-hand side of #\overline{\,\e^{\blue{z}}}#, thus finishing the proof.
Important difference between real and complex exponentials
When we deal with real numbers, let's call them #\blue{p}# and #\blue{q}#, we know that if #\e^{\blue{p}}=\e^{\blue{q}}#, we can conclude that #\blue{p}=\blue{q}#. This is not true for complex numbers. For example, consider #\blue{z}=\orange{\pi}\cdot\complexi# and #\blue{w}=\orange{3\cdot\pi}\cdot\complexi#. We have \[\e^{\orange{\pi}\cdot\complexi}=\cos\left(\orange{\pi}\right)+\sin\left(\orange{\pi}\right)\cdot\complexi = -1\]
and \[\e^{\orange{3\cdot\pi}\cdot\complexi}=\cos\left(\orange{3\cdot \pi}\right)+\sin\left(\orange{3\cdot \pi}\right)\cdot\complexi = -1\] Despite #\e^{\orange{\pi}\cdot\complexi}=\e^{\orange{3\cdot\pi}\cdot\complexi}#, we know that #\orange{\pi} \neq \orange{3\pi}#.
So far, we have only defined the exponentiation of complex numbers for base #\e#. Below, we extend this definition to other real bases.
Exponentials with other real bases
Let #t# be a positive real number and #\blue{z}=\green{a}+\orange{b}\cdot\complexi# an arbitrary complex number.
We define the exponentiation of complex numbers with base #t# as follows: \[t^{\blue{z}}=\e^{\ln\left(t\right)\cdot \blue{z}}=\e^{\ln(t)\cdot \green{a}}\cdot\left(\cos\left(\ln\left(t\right)\cdot \orange{b}\right)+\sin\left(\ln\left(t\right)\cdot \orange{b}\right)\cdot\complexi \right)\]
Note
All the properties listed in the "rules for computing with complex exponents" box also apply to exponents with different bases. If the exponent has base #t#, then the same rules listed there apply with #\e# replaced by #t#.
Consider #\blue{z}=\green{\frac{1}{2}}+\orange{4}\cdot \ii#. \[\begin{array}{rcl} 9^{\blue{z}}&=&\e^{\ln\left(9\right)\cdot\blue{z}}\\ &=&\e^{\ln(9)\cdot\green{\frac{1}{2}}}\cdot\left(\cos\left(\ln(9)\cdot\orange{4}\right)+\sin\left(\ln(9)\cdot\orange{4}\right)\cdot\ii\right)\\ &=&\e^{\ln(3)}\cdot\left(\cos\left(\ln(3)\cdot 8\right)+\sin\left(\ln(3)\cdot 8\right)\cdot \ii\right)\end{array}\] where in the last line we used #\ln\left(9\right)=2\cdot \ln(3)#.
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The complex exponential function
