Unlock the Power of Trigonometry: De Moivre’s Theorem Revealed!

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\[(\cos\theta+i\sin\theta)^{n}=\cos n\theta+i\sin n\theta,\quad(\theta\in\mathbb{ R})\]

which is a compact expression for a family of trigonometric identities. For example, in view of the binomial theorem one can show:

\[\cos 5\theta=\cos^{5}\theta-10\cos^{3}\theta\sin^{2}\theta+5\cos\theta\sin^{4 }\theta.\]

Also for \(z=re^{i\theta}\neq 0\), we have \(\frac{1}{z}=r^{-1}e^{-i\theta}\).

**Example 97**.: If \(z=1+i\), then \(\left|z\right|=\sqrt{2}\) and \(z\) lies in the first quadrant on the line \(y=x\). Hence \(z=\sqrt{2}\,e^{i\,\frac{\pi}{4}}\) and \(\frac{1}{z}=\frac{1}{\sqrt{2}}e^{-i\,\frac{\pi}{4}}\).

The fundamental theorem of algebra above guarantees that a polynomial of degree \(n\) with coefficients in \(\mathbb{C}\) has \(n\) roots (not necessarily distinct, of course). Let us consider the equation \(z^{n}=1\); using polar form of \(z=re^{i\theta}\), we must have:

\[z^{n}=1\Longleftrightarrow r^{n}e^{ni\theta}=1\Longleftrightarrow r=1,\quad \cos n\theta+i\sin n\theta=1.\]

Thus the distinct roots of the equation \(z^{n}=1\) are given by

\[e^{\frac{2k\pi i}{n}}\quad\text{for}\quad k=0,1,\dots,n-1.\]

These numbers are known as the \(n\)th _roots of unity_. Observe that the roots of \(z^{n}=1\) lie at the vertices of a regular \(n\)-gon centered at \(0\) with one vertex at \(1\) (Figure 3.12).

The first picture on the left is the cube roots of unity and

\[z^{3}=1\Longleftrightarrow z=\omega^{3}=1,\quad\text{or}\quad z=\omega=e\frac{2 \pi i}{3}\quad\text{or}\quad z=\omega^{2}=e\frac{4\pi i}{3}.\]

The picture on the right represents the sixth roots of unity.

Two special cases are worth mentioning. We have:

\[z^{4}=1\Longleftrightarrow(z^{2}-1)(z^{2}+1)=0\Longleftrightarrow z=\pm 1 \quad\text{or}\quad z=\pm i\]

and the geometric identity

\[(1-z)(1+z+z^{2}+\cdots+z^{k})=1-z^{k+1}\]

is valid for all \(z\in\mathbb{C}\) and for all natural numbers \(k\). For example, if we set \(k=2\), we obtain

\[(1-z)(1+z+z^{2})=1-z^{3},\]

which gives the cube roots of unity which we calculated before. Therefore, the roots of the equation \(1+z+z^{2}=0\) are exactly the non-real cube roots of unity, that is \(\omega\) and \(\omega^{2}\).

##### Inequalities

The following are some important inequalities that link modulus and addition.

**Proposition 40**.: _For all \(z\) and \(w\) in \(\mathbb{C}\),_

* \(|\text{Re}\,z|\leq|z|\) _and_ \(|\text{Im}\,z|\leq|z|\)_._
* \(|z+w|\leq|z|+|w|\) _(the triangle inequality)._
* \(|\,|z|-|w|\,|\leq|z+w|\)_._

Proof

: Part a) is clear since \(|z|^{2}=(\text{Re}\,z)^{2}+(\text{Im}\,z)^{2}\) and \(|z|\geq 0\). We leave it to the reader to prove part b). For c) all we need is that the following two inequalities hold:

\[|w|-|z|\leq|z+w|\quad\text{and}\quad|z|-|w|\leq|z+w|.\]

Figure 3.12: Left: cube roots of unity, Right: sixth roots of unity

Using the triangle inequality we have

\[|w|=|z+w-z|\leq|z+w|+|-z|\]

and

\[|z|=|z+w-w|\leq|z+w|+|-w|,\]

so c) follows.

**Remark.** All of the inequalities given above are between real numbers (\(\operatorname{Re}z\), \(|z|\) etc.). Note that there is no meaning assigned to an inequality \(z\leq w\). In fact it is not possible to define an ordering on \(\mathbb{C}\). Suppose such an ordering exists, then

\[i\geq 0\quad\text{implies}\quad(i)(i)\geq 0,\quad-1\geq 0\]

or

\[i\leq 0\quad\text{implies}\quad(-i)(-i)\geq 0,\quad-1\geq 0,\]

contradicting the order axioms.

### The Riemann Sphere

The extended complex plane, which consists of \(\mathbb{C}\) and the point at \(\infty\), has a convenient interpretation which we now explain. Consider the Euclidean space \(\mathbb{R}^{3}\) with coordinates \((x,y,z)\) where the \(xy\)-plane is identified with \(\mathbb{C}\). We denote by \(\mathbb{S}\) the sphere centered at \((0,0,1/2)\) of radius \(1/2\). In other words,

\[\mathbb{S}=\{(x,y,u):\quad x^{2}+y^{2}+(u-1/2)^{2}=1/4\}.\]

This sphere is the _Riemann sphere_ which touches the plane \(\mathbb{C}\) at the point \((0,0,0)\). Let \(N=(0,0,1)\) be the north pole of the sphere. Given any point \(z^{\prime}=(x,y,u)\) on \(\mathbb{S}\) different from the north pole, the line joining \(N\) to \(z^{\prime}\) intersects the \(xy\)-plane at a single point denoted by \(z=x+iy\). We call \(z\) the _stereographic projection_ of \(z^{\prime}\) (Figure 3.13).

Figure 3.13: Stereographic projection

Conversely, given any point \(z\in\mathbb{C}\), the line joining \(N\) to \(z\) intersects the sphere at \(N\) and another point which we call \(z^{\prime}\). Thus, there is a bijective correspondence between points of \(\mathbb{S}\setminus N\) and the complex plane \(\mathbb{C}\). The problem of the north pole being left out can be handled by adding an extra point \(\infty\) to \(\mathbb{C}\). We define the _extended complex plane_

\[\tilde{\mathbb{C}}:=\mathbb{C}\cup\infty.\]

We then have the natural correspondence between \(\tilde{\mathbb{C}}\) and \(\mathbb{S}\) by

\[z=x+iy\longleftrightarrow z^{\prime}=\left(\frac{x}{1+r^{2}},\frac{y}{1+r^{ 2}},\frac{r^{2}}{1+r^{2}}\right)\]

where \(r^{2}=x^{2}+y^{2}\) and under this bijection \(\infty\longleftrightarrow(0,0,1)=N\).

### Exercises

1. Prove that \(\ \frac{1}{i}=-i\\) and that \(\ \frac{1}{i+1}=\frac{1-i}{2}\\).
2. Let \(z_{1}=2+i\), \(z_{2}=3-2i\\) and \(\ z_{3}=-\frac{1}{2}+\frac{\sqrt{3}}{2}\,i.\) Evaluate:

a) \(|3z_{1}-4z_{2}|\), b) \({z_{1}}^{3}-3{z_{1}}^{2}+4z_{1}-8\), c) \(\frac{z_{1}}{z_{2}}\), d) \((\overline{z_{3}})^{4}\).
3. Find real numbers \(x\) and \(y\) such that

algebra formula visualization

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