Complex numbers: Complex functions
Complex exponential functions
We have already introduced the imaginary power \(e^{\,\mathrm{i}\,\varphi}\) for real \(\varphi\). It is not difficult to guess from this what the definition of \(e^z\) should be for a complex number \(z\). If \(z=x+\mathrm{i}\,y\), then naturally you want that the calculation rule \(e^{x+\mathrm{i}\,y}=e^x\cdot e^{\,\mathrm{i}\,y}\) remains in effect (Recall that we want to maintain as many properties of real functions as possible). But then the only possible definition for the complex exponential function \(e^z\), also referred to as \(\exp(z)\), is as follows:
Complex exponential function If \(z=x+\mathrm{i}\,y\) is in standard form, then we define \[\begin{aligned}e^z &=e^x\cdot e^{\;\mathrm{i}\,y }\\ &= e^x\cdot\bigl(\cos(y)+\mathrm{i}\,\sin(y)\bigr)\end{aligned}\]
In other words, for every complex number \(z\) we define the complex number \(e^z\) by \[\begin{cases} |\e^z| &= e^{\mathrm{Re}(z)}\\ \mathrm{arg}(e^z) &=\mathrm{Im}(z)\quad (\mathrm{mod\;} 2\pi)\end{cases} \] The function that maps \(z\) onto the complex number \(e^z\) is called called the complex exponential function and we also denote it as \(\exp\), so \(\exp(z)=e^z\).
If \(z\) is a real number, then: \(|e^z|=e^z\) and \(\mathrm{Arg}(z)=0\) . In this case, the complex exponential function value equal to the well-known, real power of \(e\).
Complex power with a positive real base With the complex exponential function in our hands we can also define powers with other positive real bases.
Let \(g\) be a positive real number and \(z\) a complex number, then we can write \[g^z = e^{\ln(g)\cdot z}\]
Via the real exponential function, logarithm, and trigonometric functions and a scientific calculator you can calculate function values.
The complex exponential function has many features that we can expect. We give two examples.
For any positive real number \(g\), integer \(n\), and each pair of complex numbers \(z\) and \(w\) the following holds: \[\begin{aligned}r^{z}\cdot r^{w}&=r^{z+w}\\ \left(r^z\right)^n&=r^{n\cdot z}\end{aligned}\]
The complex exponential function is equal to its own derivative, in other words: \[\frac{\dd }{\dd z}(e^z)=e^z\]
But there are also differences between the real exponential function and the complex exponential function. From the first multiplication rule we obtain the following statement:
The complex exponential function is periodic with period \(2\pi\,\mathrm{i}\).
After all, for every integer \(k\) the following statement is valid: \[e^{2\,k\,\pi\,\mathrm{i}}=\left(e^{\pi\,\mathrm{i}}\right)^{2k}=(-1)^{2k}=1\] So: \[e^{z+2\,k\,\pi\,\mathrm{i}}=e^z\cdot e^{2\,k\,\pi\,\mathrm{i}}=e^z\]
\(\phantom{x}\)
In other words, the last rule for the complex exponential function means that the equation \[e^z=1\] in \(\mathbb{C}\) has infinitely many solutions. The general solution can be written as \[z=0\quad(\mathrm{mod}\;2\,\pi\,\ii)\] Look at more examples of complex exponential equations that are exactly solvable.
First we set the absolute values equal to each other on the left-hand and right-hand side. On the left-hand side we have: \[|e^z|=e^{x}\] On the right-hand side we find: \[|-1+\ii|=\sqrt{(-1)^2+(1)^2}=\sqrt{2}\] Equating the absolute values leads to the equation \[\e^x=\sqrt{2}\] The exact solution of this equation is m.b.v. de rekenregels voor de natuurlijke logaritme and can be written as \[x=\tfrac{1}{2}\ln(2)\] Hereafter we set the arguments of the complex numbers on left-hand and right-hand side equal to each other. On the left-hand side we get: \[\mathrm{arg}(e^z)=\mathrm{Im}(z)=y\quad (\mathrm{mod\;} 2\pi)\] On the right-hand side special values of real and imaginary parts lead to \[-1+\ii=\sqrt{2}\cdot\bigl(\cos(\tfrac{3}{4}\pi)+\ii\,\sin(\tfrac{3}{4}\pi)\bigr)\] Thus \[\mathrm{Arg}(-1+\ii)=\tfrac{3}{4}\pi\] So \[y=\tfrac{3}{4}\pi+k\cdot 2\pi\qquad\text{for some integer }k\]
In summary: there are infinitely many solutions of the equation \[\e^z=-1+\ii\] but each solution can be written as \[z=\tfrac{1}{2}\ln(2)+(\tfrac{3}{4}\pi+k\cdot 2\pi)\,\ii \qquad\text{for some integer }k\] The general solution can be written as \[z=\tfrac{1}{2}\ln(2)+\tfrac{3}{4}\pi\,\ii\quad (\mathrm{mod\;} 2\,\pi\,\ii)\]
\(\phantom{x}\)
A shorter, but equivalent route starts with the writing of the right-hand side in polar form, say \(r\, e^{\ii\,\phi}\).
But if \[e^z=r\, e^{\ii\,\phi}\] then it follows that \[e^z=e^{\ln(r)+\ii\,\phi}\] So \[z=\ln(r)+\ii\,\phi\quad(\text{mod }2\,\pi\,\ii)\] In this exercise: \[-1+\ii=\sqrt{2}\,e^{\tfrac{3}{4}\pi\,\ii}\] So \[z=\tfrac{1}{2}\ln(2)+\tfrac{3}{4}\pi\,\ii\quad (\mathrm{mod\;} 2\,\pi\,\ii)\]