## Saturday, 11 February 2012

### Real identities involving complex numbers

This one requires some very basic college knowledge but is probably neglected by most students. With such deriving approach we can remember much less identities and realize more of the beauty of mathematics.

Part I - Nature of exponential function
In college level we first define the number e and the textbook will probably show that expansion of e directly, in better case we can get that expansion series with Taylor expansion, but that's not rigour enough. A more advanced and appropiate way is to show the expansion series of e itself, is exponential.

Definition. $exp(z)=\sum^{\infty}_{n=0}\frac{z^n}{n!}$ is defined as the exponential function. Note that z is used instead of x because we want complex number involving here.

Lemma. $exp(a+b)=exp(a)exp(b)$
Since we are not assuming $exp(z)$ as exponential, we shall demonstrates its properties. Note that $(exp(z))'=exp(z)$ is trivial in the expansion series, we have $(exp(z)exp(c-z))'=0$ by product rule, i.e., $exp(z)exp(c-z)$ is constant, and when z=0 the constant is exp(c). Now set z=a, c=a+b gives the expected result.

There could be more properties to be shown, but that's enough for us to state it as exponential.

Bridge between exponential function and trigonometric functions

Definition. $\cos z=\frac{e^{iz}+e^{-iz}}{2}$ and $\sin z=\frac{e^{iz}-e^{-iz}}{2}$.

Now  $|e^z|^2=e^z\overline{e^z}=e^ze^{\overline{z}}=e^{z+\overline{z}}=e^{2Re(z)}$, so $|e^z|=e^{Re (z)}$

Substitute them into the expansion series gives
$\cos z=1-\frac{z^2}{2!}+\frac{z^4}{4!}-...$ and $\sin z=z-\frac{z^3}{3!}+...$.

And again, it automatically fits the Euler Identity:
$e^z=cis z=\cos z+i\sin z$, $e^{\pi i}=-1$, and $\cos ^2 z+\sin ^2 z=1$. Following with the angle sum identites, showing that the function defined is the trigonometric function. (Note: the function equation concerning angle sum formula has unique solution of trigonometric function as above, this is beyond our discussion.

Part II - deriving trigonometric identities

This part is already included in my BAS note, but I'd want to show this one again.

Theorem. $\sin 2x=2\sin x\cos x$

Proof: Equating the expansion of $e^{2x}=e^xe^x$
$cis(2x)=(cis(x))^2=(\cos ^2 x-\sin^2 x)+2i\sin x\cos x$
Equating real and imaginary part,
$\sin 2x=2\sin x\cos x$.

Theorem. $\cos x\cos y=\frac{1}{2}(\cos(x+y)+\cos(x-y))$

Proof. $\cos x+\cos y=\frac{1}{4}(e^{ix}+e^{-ix})(e^{iy}+e^{-iy})$
$=\frac{1}{4}[(e^{ix+iy}+e^{-ix-iy})+(e^{ix-iy}+e^{-ix+iy})]=\frac{1}{2}(\cos (x+y)+\cos (x-y))$

There're another trick dealing with trigonometric functions at higher power.

Example. Evaluate $\int \cos ^3 x\sin ^4 x dx$
In most textbook we will try to use reduction formula, or transform it into forms of $\int (1-\sin ^2 x)\sin ^4 x d\sin x$ but we can also change it into trigoonometric function of degree 1 to deal with it.

Let $x=cis(u)$, then $x^{-1}=cis(-u)=\cos u-i\sin u$. Similarly, $x^n=cis(nu)$, $x^{-n}=\cos nu-i\sin nu$.
Observe that $x^n+x^{-n}=2\cos nu$ and $x^n-x^{-n}=2i\sin nu$

Now $\cos ^3 x\sin ^4 x=\frac{1}{128}(u+u^{-1})^3(u-u^{-1})^4=\frac{1}{128}(u^2-u^{-2})^3(u-u^{-1})$
$=\frac{1}{128}(u^6+3u^2+3u^{-2}+u^{-6})(u-u^{-1})=\frac{1}{128}(u^7-u^{-7})-(u^5+u^{-5})-3(u^3+u^{-3})+(u+u^{-1})$
$=\frac{1}{64}(\cos 7x-\cos 5x-3\cos 3x+3\cos x)$
By such a transformation the integration could be done easily. Note that $(2i)^4=2^4$.

Part III - transformation of complex number into polar (cis) form

Definition. Polar form of complex number is in form of $z=|z|cis(\theta)$ where $z_1z_2=|z_1||z_2|cis(\theta_1+\theta_2)$ The second part of the statement is called De Moivre's theorem.

The proof is simple considering the multiplication in forms of $\cos x+i\sin x$, and by angle sum formula.

Example. Turn $\sin x+i\cos x$ into polar form.

Solution.
$\sin x+i\cos x=i(\cos x-i\sin x)=cis(\frac{\pi}{4})cis(-x)=cis(\frac{\pi}{4}-x)$

Example. Turn $\frac{1-cis(x)}{1+cis(x)}$ into polar form.

Solution.
$\frac{1-\cos x-i\sin x}{1+\cos x+i\sin x}=\frac{2\sin^2 \frac{x}{2}-i\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}+i\sin\frac{x}{2}\cos\frac{x}{2}}$
$=\frac{\sin\frac{x}{2}(\sin\frac{x}{2}-i\cos\frac{x}{2})}{\cos\frac{x}{2}(\cos\frac{x}{2}+i\sin\frac{x}{2})}$
$=-i\tan\frac{x}{2}=\tan\frac{x}{2}cis(\frac{-\pi}{2})$.

Part IV - analogy on real polynomial identities
Example. Show that $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$.
In fact this is a lemma from the theorem that "any prime can be resolved into sum of two perfect squares."
It can be done by direct expansion, but we can make it by complex number:
Let $z_1=a+bi$, $z_2=c+di$
$|z_1||z_2|=(a^2+b^2)(c^2+d^2)=|z_1z_2|=(ac-bd)^2+(ad+bc)^2$, done.

Example. Show that $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$

Note that $cis(\pm\frac{2n\pi}{7})$ (n=0,1,2,3, n=0 -> 1 roots) are the roots of $x^7=1$, summing the roots gives $1+2(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7})=0$ (the imaginary parts eliminates each other since $\sin -x=-\sin x$), so we get the desired result.

More to explore:
I) Expand tangent and cotangent in terms of polynomials.
II) Express $\cos x+\cos y$ and $\tan (x+y)$ by trigonometric functions of the first degree with proof by complex numbers.
III) Turn $\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}$ into polar form, and show that $(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5})^2=-i(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5})^2$
IV1) Evaluate $\sec ^2\frac{\pi}{18}+\sec ^2\frac{5\pi}{18}+\sec ^2\frac{7\pi}{18}$
IV2) In fact we have another real identities concerning sum of four squares:
$(a^2+b^2+c^2+d^2)(e^2+f^2+g^2+h^2)=(ae+bf+cg+dh)^2+(af-be+ch-dg)^2+(ag-bh-ce+df)^2+(ah+bg-cf-de)^2$ which can be proved by quaternion numbers. Search for quaternion numbers and prove the above identity by quaternion numbers.

Reference
Walter Rudin. Real and complex analysis. (In fact, the analysis on exponential series is commonly appearing in most books)
Margaret M.Gow. A Course in pure mathematics.