### 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 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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