Complex numbers: Complex functions
Rules of calculation for complex powers
In the new format of imaginary powers of #\e#, the multiplication and division of complex numbers on the unit circle looks a lot neater. The usual rules of calculation may also be used for complex powers of #\e#, and hence, also for every other positive real base:
Products of complex powers
Let #r# be a positive real number. For any two complex numbers #z# and #w#, and every integer #n# applies:
- #r^{z}\cdot r^{w}=r^{z+w}#
- #\left(r^{z}\right)^n=r^{n\cdot z}#
Proof of 1: We first derive the outcome in the special case #r=\e#. The complex numbers in the left and right hand side have the same absolute value, which can be shown by using the theory Calculating with polar coordinates and Rules of calculation for real and imaginary parts :
\[ \begin{array}{rcl}| \e^{z} \cdot \e^{w}| & =& | \e^{z}|\cdot | \e^{w}| =
\e^{{\Re} (z)}\cdot \e^{{\Re} (w)}= \e^{{\Re} (z)+{\Re} (w)}\tiny,\\ \\
| \e^{z+w}|\ &=& \e^{{\Re} (z+ w)}=\e^{{\Re} (z)+{\Re} (w)}\tiny.\end {array}
\]
They also have the same argument:
\[ \begin{array}{rcl}
\arg (\e^{z}\cdot \e^{w})&=&\arg (\e^{z})+\arg (\e^{w})={\Im} (z)+{\Im} (w)\tiny ,
\\
\arg (\e^{z+w})&=&{\Im} (z+w)\ ={\Im} (z)+{\Im} (w)\tiny .\end {array}
\]
Because the numbers on the left and right hand side have the same absolute value and the same argument, they are equal. With this we have derived #\e^{z}\cdot \e^{w}=\e^{z+w}#.
The general case follows from the following derivation: \[r^{z}\cdot r^{w}=\e^{\ln(r)\cdot z}\cdot \e^{\ln(r)\cdot w}=\e^{\ln(r)\cdot z+\ln(r)\cdot w}=\e^{\ln(r)\cdot( z+ w)}=r^{z+w}\tiny.\]
Proof of 2: For #n=1# it states #\left(r^{z}\right)^1=r^{1\cdot z}#, which is obvious because both members are equal to #r^z#.
We will continue with full induction to prove the statement for all natural numbers #n#. For this we assume that the statement is true for #n# and derive the statement also for #n+1# instead of #n#. Hence, we assume that #\left(r^{z}\right)^n=r^{n\cdot z}# is true (this statement is called the induction hypothesis), and from this we reduce that #\left(r^{z}\right)^{n+1}=r^{(n+1)\cdot z}# : \[ \begin{array}{rcl}\left(r^{z}\right)^{n+1}&=&\left(r^{z}\right)^{1}\cdot \left(r^{z}\right)^{n}\\&&\phantom{uvwxyzuvwxyz}\color{blue}{w^{n+1}=w\cdot w^n}\\&=& r^{z} \cdot r^{n\cdot z}\\&&\phantom{uvwxyzuvwxyz}\color{blue}{\text{case }n=1\text{ and induction hypothesis}}\\&=& r^{z+n\cdot z}\\&&\phantom{uvwxyzuvwxyz}\color{blue}{\text{ equal 1}}\\&=& r^{(n+1)\cdot z}\\&&\phantom{uvwxyzuvwxyz}\color{blue}{z\text{ moved outside of the brackets}}\end {array}\] according to the principle of full induction the statement is hereby proven all natural numbers #n#.
We still need to prove the statement for all integers #n\le0#. If #n=0#, we find #\left(r^{z}\right)^0=1=r^0=r^{0\cdot z}#, proving the statement for this case.
The statement #n=-1# follows from the first statement with #w=-z#: #r^{z}\cdot r^{-z}=r^{zz}=r^0=1# gives #\frac{1}{r^z}=r^{-z}# such that #\left(r^z\right)^{-1}=\frac{1}{r^z}=r^{-z}=r^{-1\cdot z}#.
Finally, for #n\lt -1#, we write #m=-n# and we use the fact that #\left(r^{z}\right)^m=r^{m\cdot z}# is already proven: \[\left(r^{z}\right)^{n}=\left(r^{z}\right)^{-m}=\frac{1}{\left(r^{z}\right)^{m}}=\frac{1}{r^{m\cdot z}}={r^{-m\cdot z}}={r^{n\cdot z}}\] with which the statement is proven for all integer #n#.
A famous consequence of these rules of calculation is the following.
de Moivre's formula
If #\varphi# is real, then for every integer #n# holds: \[\left( \cos( \varphi)+\sin(\varphi)\cdot\ii\right)^n = \cos(n\cdot\varphi)+\sin(n\cdot\varphi)\cdot\ii\tiny.\]
This is a result of the second equality #\left(\e^{z}\right)^n=\e^{n\cdot z}# from the theorem above, Products of complex powers, and two applications of Euler's Formula: \[ \begin{array}{rcl}\left(\cos(\varphi)+\sin(\varphi)\cdot\ii\right)^n &=&\left(\e^{\varphi\cdot \ii}\right)^n\\&=& \e^{n\cdot\varphi\cdot \ii}\\&=&\cos(n\cdot\varphi)+\sin(n\cdot\varphi)\cdot\ii\end {array}\]
The strength of the rule of the Moivre becomes apparent in the following determinations of all complex higher power roots of positive real numbers.
Higher Power Roots
Let #w# be a complex number not equal to #0# and #n# a natural number. The equation \[z^n=w\] with unknown #z# has exactly #n# various solutions, namely \[z=\sqrt[n]{|w|}\cdot \e^{\frac{\left(\arg(w)+2\,k\cdot\pi\right)\cdot\ii}{n}}\phantom{xxxxx}\text{ in which }k=0,1,\ldots,n-1\tiny.\]
Proof: Let #z=r\cdot \e^{\varphi\cdot\ii}# be the polar form of #z#; hence, #r# is a positive real number and #\varphi# is an arbitrary real number. Because of the rule of De Moivre #z^n=r^n\cdot\e^{n\cdot\varphi\cdot\ii}# applies. Equating to #w# and comparison of absolute value and argument gives \[\eqs{r^n&=&|w|\cr n\cdot\varphi&=&\arg(w)\pmod{2\pi}\cr}\] the first equation yields #r=|w|^{\frac{1}{n}}=\sqrt[n]{|w|}# because #r# is positive (see the theory higher power roots). The second equations teaches us that there is an integer #k# such that #n\cdot\varphi=\arg(w)+2k\cdot \pi#. Hence, #\varphi=\frac{\arg(w)+2k\cdot \pi}{n}#. Therefore \[z=r\cdot\e^{\varphi\cdot\ii}=\sqrt[n]{|w|}\cdot\e^{\frac{\left(\arg(w)+2\,k\cdot\pi\right)\cdot\ii}{n}}\tiny.\]
We will check that we can demand #k=0,1,\ldots,n-1#. If we at to #k# an integer multiple of #n#, say #q\cdot n#, the solution does not change: \[\sqrt[n]{|w|}\cdot\e^{\frac{\left(\arg(w)+2\,(k+q\cdot n)\cdot\pi\right)\cdot\ii}{n}}=\sqrt[n]{|w|}\cdot\e^{\frac{\left(\arg(w)+2\,k\cdot\pi\right)\cdot\ii}{n}}\cdot\e^{2\,q\cdot\pi\cdot\ii}=\sqrt[n]{|w|}\cdot\e^{\frac{\left(\arg(w)+2\,k\cdot\pi\right)\cdot\ii}{n}}\tiny.\] As a result we can limit ourselves to values of #k# in the interval #\ivco{0}{n}#. On the other hand, the #n# remaining values of #k# give various arguments modulo #2\pi#, and therefore different solutions . This proves the theorem.
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