It follows from the properties of real numbers that, the square of a real number is never negative. Consequently, the elementary quadratic equation $x^2+1=0$, has no solution in the system of real numbers. Introduction of a new type of numbers, called complex numbers, has made it possible to provide solutions to the equation $x^2+1=0$, and also of the more general type of equations:

$$a_0x^n+a_1x^{n-1}+\ldots+a_n=0$$

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Complex Numbers

Any complex number $z$ is an ordered pair of real numbers $(a,b)$, that satisfies the following condition (1), and the following laws of operations (2), and (3):

1. $(a,b)=(c,d)$, if and only if $a=c$, $b=d$. This is known as the condition of equality.

2. $(a,b)+(c,d)=(a+c,b+d)$. This is known as the definition of addition.

3. $(a,b).(c,d)=(ac-bd,ad+bc)$. This is known as the definition of multiplication.

Theorems for complex numbers

1. Addition of two complex numbers is commutative

Considering two complex numbers, $z_1=(a_1,b_1)$, and $z_2=(a_2,b_2)$. Then, as per the theorem, $z_1+z_2=z_2+z_1$.

$$\begin{align*} z_1+z_2&=(a_1,b_1)+(a_2,b_2)\\ &=(a_1+a_2,b_1+b_2) \end{align*}$$

But, $a_1+a_2=a_2+a_1$, and, $b_1+b_2=b_2+b_1$, since addition of real numbers is commutative.

$$\begin{align*} \therefore z_1+z_2&=(a_1+a_2,b_1+b_2)\\ &=(a_2+a_1,b_2+b_1)\\ &=(a_2,b_2)+(a_1,b_1)=z_2+z_1 \end{align*}$$

Therefore, the addition of two complex numbers is commutative:

$$z_1+z_2=z_2+z_1$$

2. Addition of complex numbers is associative.

Considering three complex numbers, $z_1=(a_1,b_1)$, $z_2=(a_2,b_2)$, and $z_3=(a_3,b_3)$. Then, as per the theorem, $z_1+(z_2+z_3)=(z_1+z_2)+z_3$.

$$\begin{align*} z_1+(z_2+z_3)&=(a_1,b_1)+\{(a_2,b_2)+(a_3,b_3)\}\\ &=(a_1,b_1)+(a_2+a_3,b_2+b_3)\\ &=(a_1+(a_2+a_3),b_1+(b_2+b_3)) \end{align*}$$

But, $a_1+(a_2+a_3)=(a_1+a_2)+a_3$, and $b_1+(b_2+b_3)=(b_1+b_2)+b_3$, since addition of real numbers is associative.

$$\begin{align*} \therefore z_1+(z_2+z_3)&=(a_1+(a_2+a_3),b_1+(b_2+b_3))\\ &=((a_1+a_2)+a_3,(b_1+b_2)+b_3)\\ &=(a_1+a_2,b_1+b_2)+(a_3,b_3)\\ &=\{(a_1,b_1)+(a_2,b_2)\}+(a_3,b_3)\\ &=(z_1+z_2)+z_3 \end{align*}$$

Therefore, the addition of complex numbers is associative:

$$z_1+(z_2+z_3)=(z_1+z_2)+z_3$$

3. Multiplication of two complex numbers is commutative

Considering two complex numbers, $z_1=(a_1,b_1)$, and $z_2=(a_2,b_2)$. Then, as per the theorem, $z_1.z_2=z_2.z_1$.

$$\begin{align*} z_1.z_2&=(a_1,b_1).(a_2,b_2)\\ &=(a_1a_2-b_1b_2,a_1b_2+b_1a_2) \end{align*}$$

But, $a_1a_2=a_2a_1$, $b_1b_2=b_2b_1$, $a_1b_2=b_2a_1$, and $b_1a_2=a_2b_1$, since the multiplication of real numbers is commutative.

$$\begin{align*} \therefore z_1.z_2&=(a_1a_2-b_1b_2,a_1b_2+b_1a_2)\\ &=(a_2a_1-b_2b_1,b_2a_1+a_2b_1) \end{align*}$$

But, $b_2a_1+a_2b_1=a_2b_1+b_2a_1$, since addition of real numbers is commutative.

$$\begin{align*} \therefore z_1.z_2&=(a_2a_1-b_2b_1,b_2a_1+a_2b_1)\\ &=(a_2a_1-b_2b_1,a_2b_1+b_2a_1)\\ &=(a_2,b_2).(a_1,b_1)=z_2.z_1 \end{align*}$$

Therefore, the multiplication of two complex numbers is commutative.

$$z_1.z_2=z_2.z_1$$

4. Multiplication of complex numbers is associative

Considering three complex numbers, $z_1=(a_1,b_1)$, $z_2=(a_2,b_2)$, and $z_3=(a_3,b_3)$. Then, as per the theorem, $z_1.(z_2.z_3)=(z_1.z_2).z_3$.

$$\begin{align*} z_1.(z_2.z_3)&=(a_1,b_1).\{(a_2,b_2).(a_3,b_3)\}\\ &=(a_1,b_1)(a_2a_3-b_2b_3,a_2b_3+b_2a_3)\\ &=(a_1(a_2a_3-b_2b_3)-b_1(a_2b_3+b_2a_3),a_1(a_2b_3+b_2a_3)+b_1(a_2a_3-b_2b_3))\\ &=(a_1a_2a_3-a_1b_2b_3-b_1a_2b_3-b_1b_2a_3,a_1a_2b_3+a_1b_2a_3+b_1a_2a_3-b_1b_2b_3)\\ &=((a_1a_2-b_1b_2)a_3-(a_1b_2+b_1a_2)b_3,(a_1a_2-b_1b_2)b_3+(a_1b_2+b_1a_2)a_3)\\ &=(a_1a_2-b_1b_2,a_1b_2+b_1a_2).(a_3,b_3)=\{(a_1,b_1).(a_2,b_2)\}.(a_3,b_3)\\ &=(z_1.z_2).z_3 \end{align*}$$

Therefore, multiplication of complex numbers is associative.

$$z_1.(z_2.z_3)=(z_1.z_2).z_3$$

Unit complex number

Complex numbers just like real numbers, when multipled with one, give the complex number itself. For example, considering a complex number, $z_1=(a_1,b_1)$, as per the above statement, $z_1.1=z_1$.

$$\begin{align*} z_1.1&=(a_1,b_1).(1,0)\\ &=(a_1.1-b_1.0,a_1.0+b_1.1)\\ &=(a_1,b_1)=z_1 \end{align*}$$ $$\therefore (a,b).(1,0)=(a,b)$$

Here, $(1,0)$, is said to be unit complex number, denoted by $1$.

Negative of a complex number

Negative of a complex number $z_1$, is denoted by $-z_1$, is a complex number $z_2$, such that, $z_1+z_2=0$.

Considering $z_1=(a_1,b_1)$, and $z_2=(a_2,b_2)$,

$$\begin{align*} z_1+z_2=0\Rightarrow (a_1,b_1)+(a_2,b_2)=(0,0)\\ \Rightarrow (a_1+a_2,b_1+b_2)=(0,0) \end{align*}$$

From the above equation, we have, $a_1+a_2=0\Rightarrow a_2=-a_1$, and $b_1+b_2=0\Rightarrow b_2=-b_1$

$$\therefore -z_1=z_2=(a_2,b_2)=(-a_1,-b_1)=-(a_1,b_1)$$

Thus, the negative of a complex number $z_1=(a,b)$, is $-z_1=-(a,b)$.

Subtraction of a complex number

Let two complex numbers be, $z_1=(a_1,b_1)$, and $z_2=(a_2,b_2)$. Their subtraction $z_1-z_2$, is:

$$\begin{align*} z_1-z_2&=z_1+(-z_2)\\ &=(a_1,b_1)+(-a_2,-b_2)\\ &=(a_1-a_2,b_1-b_2) \end{align*}$$

Therefore, $z_1-z_2=(a_1-a_2,b_1-b_2)$.

Multiplicative inverse of a complex number

Multiplicative inverse, of any non-zero complex number $z_1$, which is denoted as $z_1^{-1}$, is a complex number $z_2$, so that, $z_1.z_2=1$

Considering two complex numbers, $z_1=(a_1,b_1)$, and $z_2=(a_2,b_2)$.

$$\begin{align*} z_1.z_2=1&\Rightarrow (a_1,b_1).(a_2,b_2)=(1,0)\\ &\Rightarrow (a_1a_2-b_1b_2,a_1b_2+b_1a_2)=(1,0) \end{align*}$$

From above, we have two equations,

$$\begin{equation} a_1a_2-b_1b_2=1\Rightarrow b_2=\frac{a_1a_2-1}{b_1} \end{equation}$$ $$\begin{equation} a_1b_2+b_1a_2=0\Rightarrow b_2=-\frac{b_1a_2}{a_1} \end{equation}$$

Comparing the two equations, we get,

$$\begin{align*} \frac{a_1a_2-1}{b_1}=-\frac{b_1a_2}{a_1}&\Rightarrow a_1^2a_2-a_1=-b_1^2a_2\\ &\Rightarrow (a_1^2+b_1^2)a_2=a_1\\ &\Rightarrow a_2=\frac{a_1}{a_1^2+b_1^2} \end{align*}$$

Putting the value of $a_2$ in $(2)$, we get:

$$b_2=-\frac{b_1}{a_1}\frac{a_1}{a_1^2+b_1^2}=\frac{-b_1}{a_1^2+b_1^2}$$

Now,

$$\because z_1^{-1}=z_2=\left(\frac{a_1}{a_1^2+b_1^2},\frac{-b_1}{a_1^2+b_1^2}\right)\\$$ $$\begin{equation} \therefore z_1^{-1}=\left(\frac{a_1}{a_1^2+b_1^2},\frac{-b_1}{a_1^2+b_1^2}\right) \end{equation}$$

Division of a complex number

Considering two complex numbers, $z_1=(a_1,b_1)$, and $z_2=(a_2,b_2)$, where $z_2$ is a non-zero complex number.

$$\begin{align*} \therefore \frac{z_1}{z_2}&=z_1.z_2^{-1}=(a_1,b_1).(a_2,b_2)^{-1}\\ &=(a_1,b_1).\left(\frac{a_2}{a_2^2+b_2^2},\frac{-b_2}{a_2^2+b_2^2}\right)\\ &=\left(\frac{a_1a_2}{a_2^2+b_2^2}-\frac{b_1b_2}{a_2^2+b_2^2},\frac{-a_1b_2}{a_2^2+b_2^2}+\frac{b_1a_2}{a_2^2+b_2^2}\right)\\ &=\left(\frac{a_1a_2-b_1b_2}{a_2^2+b_2^2},\frac{-a_1b_2+b_1a_2}{a_2^2+b_2^2}\right) \end{align*}$$ $$\therefore \frac{z_1}{z_2}=\left(\frac{a_1a_2-b_1b_2}{a_2^2+b_2^2},\frac{-a_1b_2+b_1a_2}{a_2^2+b_2^2}\right)$$

If the product of two complex numbers is zero, then atleast one of them is zero.

Considering two complex numbers, $z_1=(a_1,b_1)$, $z_2=(a_2,b_2)$, such that $z_1.z_2=0$.

$$\begin{align*} z_1.z_2=0&\Rightarrow(a_1,b_1).(a_2,b_2)=(0,0)\\ &\Rightarrow (a_1a_2-b_1b_2,a_1b_2+b_1a_2)=(0,0)\\ \end{align*}$$

We have two equations:

$$\begin{equation} a_1a_2-b_1b_2=0 \end{equation}$$ $$\begin{equation} a_1b_2+b_1a_2=0 \end{equation}$$

Let, $z_2\ne0$, then, $a_2^2+b_2^2\ne0$

Multiplying $(4)$ by $a_2$, and $(5)$ by $b_2$, and adding:

$$\begin{align*} &\Rightarrow a_2(a_1a_2-b_1b_2)+b_2(a_1b_2+b_1a_2)=0\\ &\Rightarrow a_1a_2^2-b_1b_2a_1+a_1b_2^2+b_1a_2b_2=0\\ &\Rightarrow a_1(a_2^2+b_2^2)=0 \end{align*}$$

Similarly multiplying $(4)$ by $b_2$, and $(5)$ by $a_2$, and subtracting, we have:

$$b_1(a_2^2+b_2^2)=0$$

Therefore, from above, we have: $a_1=b_1=0$. Thus, $z_1=0$, for $z_2\ne0$

Similarly, for $z_1\ne0$, we have $z_2=0$

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Normal Form

A complex number, $z=(a,b)$, in the normal form, can be written as, $z=a+bi$.

Addition in normal form

Let $z_1=(a,b)$, and, $z_2=(c,d)$. In normal form they can be written as, $z_1=a+ib$, and $z_2=c+id$.

$$\begin{align*} z_1+z_2&=(a+ib)+(c+id)\\ &=(a+c)+i(b+d) \end{align*}$$

Multiplication in normal form

Let $z_1=a+ib$, and $z_2=c+id$. Therefore their multiplication will be:

$$\begin{align*} z_1.z_2&=(a+ib).(c+id)\\ &=(ac-bd)+i(ad+bc) \end{align*}$$

Multiplication can also be performed as real binomials, provided that $i^2=-1$. For example, considering two complex numbers, $z_1=a+ib$, and $z_2=c+id$.

$$\begin{align*} z_1.z_2&=(a+ib).(c+id)\\ &=ac+i.ad+i.bc+i^2.bd\\ &=ac-bd+i(ad+bc) \end{align*}$$

Example 1: Find out the product of two complex numbers $2+3i$, and $3+i$.

$$\begin{align*} (2+3i).(3+i)&=(2.3-3.1)+(2.1+3.3)i\\ &=3+11i \end{align*}$$

Example 2: Find out the result of division of two complex numbers $5+9i$ by $2+11i$.

$$\begin{align*} \frac{5+9i}{2+11i}&=(5+9i).(2+11i)^{-1}\\ &=(5+9i)\left(\frac{2}{4+121}+\frac{-11}{4+121}i\right)\\ &=\left(\frac{10+99}{125}+\frac{-55+18}{125}i\right)\\ &=\left(\frac{109}{125}-\frac{37}{125}i\right)\\ &=\frac{109-37i}{125} \end{align*}$$

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Polar Form

Considering a complex number: $z=a+ib$. When represented in the complex plane, $z$ is represented by the point whose Cartesian co-ordinates are $(a,b)$, referred to two perpendicular lines as axes, the first co-ordinate axis being called the real axis, and the second the imaginary axis.

Taking the origin as the pole and the real axis as the initial line, let $(r,\theta)$, as the polar coordinates of the point $(a,b)$. Then it can be concurred that, $a=r\cos{\theta}$, and, $b=r\sin\theta$. Geometrically, $r$ is said to be the distance between the points $(a,b)$, from origin. And this $r$ is the modulus of the complex number $z$.

We know: $a=r\cos\theta$, and $b=r\sin\theta$.

$$\begin{align*} &\therefore r\cos\theta+r\sin\theta=a+b\\ &\Rightarrow r^2\cos^2\theta+r^2\sin^2\theta=a^2+b^2\\ &\Rightarrow r^2(\cos^2\theta+\sin^2\theta)=a^2+b^2 \end{align*}$$

We know: $\sin^2\theta+\cos^2\theta=1$

$$\therefore r^2(\cos^2\theta+\sin^2\theta)=a^2+b^2\\ \Rightarrow r^2=a^2+b^2\Rightarrow r=\sqrt{a^2+b^2}$$

Therefore, the modulus of a complex number $z$:

$$|z|=r=\sqrt{a^2+b^2}$$

The argument of a polar form represented as $\textnormal{Arg}\ z$, are all values of $\theta$, satisfying the relations $\cos\theta=\frac ar$, and $\sin\theta=\frac br$. Let's say, $\alpha$ is a value of $\theta$, satisfying the above relation, then $\textnormal{Arg}\ z=\alpha+2n\pi$, where $n$ is an integer.

The principal argument of $z$, also called the amplitude of $z$, denoted by $\textnormal{Arg}\ z$, is defined to be the angle $\theta$ which satisfies the ineqality $-\pi\lt\theta\le\pi$. For example, lets say, $\theta=\frac{3\pi}{2}$, is an argument of $z$, but not the principle argument, because it doesnt lie within the given inequality: $-\pi\lt\theta\le\pi$.

Principle argument is also an argument. but within the argument set. Furthermore, the arguments cannot be termined from $\tan^-1{\frac ba}$.

Example 3: Given a complex number: $z=-1+i$, find the $\textnormal{mod.}\ z$, $\textnormal{arg.}\ z$, and express $z$ in polar form.

Given, $z=-1+i$. Comparing it with $z=r\cos\theta+r\sin\theta$, we have:

$$r\cos\theta=-1\\ r\sin\theta=1$$

Modulus of $z$:

$$|z|=r=\sqrt{a^2+b^2}=\sqrt{-1^2+1^2}=\sqrt2$$

Argument of $z$:

$$\textnormal{Arg. }z=\theta=\cos^{-1}\left({\frac {-1}{\sqrt2}}\right)=\frac{3\pi}4$$

$z$ in polar form:

$$z=\sqrt2\left(\cos\frac{3\pi}{4}+\sin\frac{3\pi}4i\right)$$

Example 3: Given a complex number: $z=-1-i$, find the $\textnormal{mod.}\ z$, $\textnormal{arg.}\ z$, and express $z$ in polar form.

Given, $z=-1-i$. Comparing it with $z=r\cos\theta+r\sin\theta$, we have:

$$r\cos\theta=-1\\ r\sin\theta=-1$$

Modulus of $z$:

$$|z|=r=\sqrt{a^2+b^2}=\sqrt{(-1)^2+(-1)^2}=\sqrt2$$

Argument of $z$:

$$\textnormal{arg. }z=\cos^{-1}\left(-{1\over r}\right)=\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3\pi}{4},\frac{5\pi}{4}\ldots\\ \textnormal{arg. }z=\sin^{-1}\left(-{1\over r}\right)=\sin^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{5\pi}{4}\ldots$$ $$\therefore \textnormal{arg. }z=\frac{5\pi}{4}$$

De Morgan's Theorem

If $n$ be any integer then:

$$\begin{align} (\cos{\theta}+i\sin{\theta})^n=\cos{n\theta}+i\sin n\theta \end{align}$$

If $n$ be a fraction, then one of the values of $(\cos\theta+i\sin\theta)^n$ is:

$$\begin{align} (\cos\theta+i\sin\theta)^n=(\cos n\theta+i\sin n\theta) \end{align}$$

Case 1: $n\in\N$

For, $n=1$:

$$\begin{align*} (\cos\theta+i\sin\theta)^1&=\cos\theta+i\sin\theta\\ &=\cos1.\theta+i\sin1.\theta \end{align*}$$

Let $n=k$,

$$\begin{align*} (cos\theta+i\sin\theta)^k=\cos k\theta+i\sin k\theta \end{align*}$$ $$\begin{align*} \therefore (\cos\theta+i\sin\theta)^{k+1}&=(\cos k\theta+i\sin k\theta)(\cos k\theta+i\sin k\theta)\\ &=(\cos k\theta\cos\theta-\sin k\theta\sin\theta)+i(\sin k\theta\cos\theta+\cos k\theta\sin\theta)\\ &=\cos(k+1)\theta+i\sin(k+1)\theta \end{align*}$$ $$\begin{align} \therefore (\cos\theta+i\sin\theta)^{k+1}=\cos(k+1)\theta+i\sin(k+1)\theta \end{align}$$

Case 2: $n=-m, n\in\N$

$$\begin{align*} (\cos\theta+i\sin\theta)^n&=(\cos\theta+i\sin\theta)^{-m}\\ &=\frac{1}{(\cos\theta+i\sin\theta)^m}\\ &=\frac{1}{\cos m\theta+i\sin m\theta}\\ &=\frac{\cos m\theta-i\sin m\theta}{1}&&[\text{Conjugate multiplication}]\\ &=\cos(-m)\theta+i\sin(-m)\theta\\ &=\cos n\theta+i\sin n\theta&&[\because n=-m] \end{align*}$$

Therefore, the theorem holds true for all negative numbers.

Case 3: $n=\frac{p}{q}, \{p,q\}\in\Z, q>0$

$$\begin{align*} \left(\cos\frac{p}{q}\theta+i\sin\frac{p}{q}\theta\right)^q&=\cos p.\theta+i\sin p.\theta\\ &=(\cos\theta+i\sin\theta)^p \end{align*}$$

Therefore, one of the values of $(\cos\theta+i\sin\theta)^{p/q}$ is:

$$\begin{align} (\cos\theta+i\sin\theta)^{\frac{p}{q}}=\left(\cos\frac{p}{q}\theta+i\sin\frac{p}{q}\theta\right) \end{align}$$

Therefore, one of the values of $(\cos\theta+i\sin\theta)^n$ is $(\cos n\theta+i\sin n\theta)$.

n-th Root of a Complex Number

Considering a complex number: $z=r(\cos\theta+i\sin\theta)$, where $z\in C$. In the above $r=|z|$, which is known as the modulus of complex number $z$. $\theta$ is called the argument of $z$, and written as $\text{arg}(z)$, and its value lies between:

$$\begin{align} -\pi<\text{arg}(z)\le\pi \end{align}$$

Considering the following:

$$\begin{align*} &\ z=r[\cos(\theta+2k\pi)+i\sin(\theta+2k\pi)],&&k\in\Z\\ \Rightarrow\ &z^{1/n}=r^{1/n}[\cos(\theta+2k\pi)+i\sin(\theta+2k\pi)]^{1/n}\\ \Rightarrow\ &z^{1/n}=r^{1/n}\left[\cos\left(\frac{\theta+2k\pi}{n}\right)+i\sin\left(\frac{\theta+2k\pi}{n}\right)\right] \end{align*}$$

Therefore, distinct values of $z^{1/n}$ are obtained for:

$$k=0,1,2,3,\ldots,n-1$$

Example 4: Find out the cube roots of unity.

Given,

$$\begin{align*} z&=1\\ &=1(\cos0+i\sin0)\\ &=\cos{2k\pi}+i\sin{2k\pi},k\in\Z&&[\because \sin\theta=\sin({2k\pi+\theta})] \end{align*}$$

Finding out the cube root:

$$\begin{align*} \therefore z^{1/3}&=(\cos{2k\pi}+i\sin{2k\pi})^{1/3},&&k=0,1,2\\ &=\cos{\frac{2k\pi}{3}}+i\sin{\frac{2k\pi}{3}},&&k=0,1,2 \end{align*}$$

Substituting the value of $z=1$, and $k=0,1,2$, we have:

$$\begin{align*} 1^{1/3}&=\cos0+i\sin0,\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}},\cos{\frac{4\pi}{3}}+i\sin{\frac{4\pi}{3}}\\ &=1,-\frac{1}{2}+i\frac{\sqrt{3}}{2},-\frac{1}{2}-i\frac{\sqrt{3}}{2}\\ &=1,\frac{-1\pm i\sqrt{3}}{2} \end{align*}$$

Expansions of Cosine and Sine

$$\begin{align*} \cos{n\theta}+i\sin{n\theta}&=(\cos\theta+i\sin\theta)^n\\ &=\cos^n\theta+{}^nC_1.\cos^{n-1}\theta.(i\sin\theta)+{}^nC_2.\cos^{n-2}\theta.(i\sin\theta)^2+{}^nC_3.\cos^{n-3}\theta.(i\sin\theta)^3+\ldots\\ &=\cos^n\theta+i.{}^nC_1\cos^{n-1}\sin\theta-{}^nC_2\cos^{n-2}\theta.\sin^2\theta-i.{}^nC_3\cos^{n-3}\theta.\sin^3\theta+{}^nC_4\cos^{n-4}\theta.\sin^4\theta+\ldots\\ &=(cos^n\theta-{}^nC_2\cos^{n-2}\theta\sin^2\theta+{}^nC_4\cos^{n-4}\theta.sin^4\theta-\ldots)+i({}^nC_1\cos^{n-1}\theta.sin\theta-{}^nC_3\cos^{n-3}\theta.\sin^3\theta+\ldots) \end{align*}$$

From the above, we get:

$$\begin{align} \cos{n\theta}=cos^n\theta-{}^nC_2\cos^{n-2}\theta\sin^2\theta+{}^nC_4\cos^{n-4}\theta.sin^4\theta-\ldots\\ \sin{n\theta}={}^nC_1\cos^{n-1}\theta.sin\theta-{}^nC_3\cos^{n-3}\theta.\sin^3\theta+{}^nC_5\cos^{n-5}\theta.\sin^5\theta-\ldots \end{align}$$

Expansion of Trigonometric Expressions in Power Series

Trigonometric expressions such as $\sin\alpha$, $\cos\alpha$ can be expanded in the power series of $\alpha$. Considering $\cos{n\theta}$:

$$\begin{align*} \cos{n\theta}&=cos^n\theta-{}^nC_2\cos^{n-2}\theta\sin^2\theta+{}^nC_4\cos^{n-4}\theta.sin^4\theta-\ldots\\ &=\cos^n\theta-\frac{n(n-1)}{2!}\cos^{n-2}\theta.sin^2\theta+\frac{n(n-1)(n-2)(n-3)}{4!}\cos^{n-4}\theta.\sin^4\theta-\ldots \end{align*}$$

Now, $n\theta=\alpha\Rightarrow n=\frac{\alpha}{\theta}$

$$\begin{align*} \therefore \cos\alpha=\cos^n\theta&-\frac{\frac{\alpha}{\theta}\left(\frac{\alpha}{\theta}-1\right)}{2!}\cos^{n-2}\theta\left(\frac{\sin\theta}{\theta}\right)^2\\ &+\frac{\frac{\alpha}{\theta}\left(\frac{\alpha}{\theta}-1\right)\left(\frac{\alpha}{\theta}-2\right)\left(\frac{\alpha}{\theta}-3\right)}{4!}\cos^{n-4}\theta\left(\frac{\sin\theta}{\theta}\right)^4-\ldots \end{align*}$$

Now, $n\to\infty, \theta\to0$, such that $\alpha$ is constant, we get:

$$\begin{align} \cos\alpha=1-\frac{\alpha^2}{2!}+\frac{\alpha^4}{4!}-\ldots\\ \sin\alpha=\alpha-\frac{\alpha^3}{3!}+\frac{\alpha^5}{5!}-\ldots \end{align}$$

We get:

$$\begin{align*} \tan\alpha&=\frac{\sin\alpha}{\cos\alpha}=\frac{\alpha-\frac{\alpha^3}{3!}+\frac{\alpha^5}{5!}-\ldots}{1-\frac{\alpha^2}{2!}+\frac{\alpha^4}{4!}-\ldots}\\ &=\left(\alpha-\frac{\alpha^3}{3!}+\frac{\alpha^5}{5!}-\ldots\right)\left\{1-\left(\frac{\alpha^2}{2!}-\frac{\alpha^4}{4!}+\ldots\right)\right\}^{-1}\\ &=\left(\alpha-\frac{\alpha^3}{3!}+\frac{\alpha^5}{5!}-\ldots\right)\left\{1-\left(\frac{\alpha^2}{2!}-\frac{\alpha^4}{4!}+\ldots\right)+\left(\frac{\alpha^2}{2!}-\frac{\alpha^4}{4!}+\ldots\right)^2+\ldots\right\}\\ &=\alpha+\frac{1}{3}\alpha^3+\frac{2}{15}\alpha^5+\ldots+\infin \end{align*}$$

Therefore, we get:

$$\begin{align} \tan\alpha=\alpha+\frac{1}{3}\alpha^3+\frac{2}{15}\alpha^5+\ldots+\infin \end{align}$$