More on Complex Numbers - MathsTips.com

The earlier chapter on Complex Numbers explained what we mean by a complex number together with a few properties of complex numbers. Basic mathematical operations of complex numbers were introduced. So, the previous chapter may be considered as the preliminary base on which this chapter aims to build further. In this chapter we shall cover a few advanced topics under complex numbers.

Amplitude(or Argument) of a complex number:

Let $z=x+iy$ where $x,y$ are real, $i=\sqrt{-1}$ and $x^2+y^2\neq 0$ ; then the value of $\theta$ for which the equations:

$x=\lvert z \rvert \cos\theta$ …(1) and

$y=\lvert z \rvert \sin\theta$ …(2)

are simultaneously satisfied is called the Argument(or Amplitude) of $z$ and is denoted by $Arg z (or, \: Ampz)$ .

Clearly, equations (1) and (2) are satisfied for infinite values of $\theta$ ; any of these values of $\theta$ is the value of $Amp z$ . However, the unique value of $\theta$ lying in the interval $-\pi < \theta \leq \pi$ and satisfying equations (1) and (2) is called the principal value of $Arg z$ and we denote this principal value by $arg z$ or $amp z$ .Unless otherwise mentioned, by argument of a complex number we mean its principal value.

Since, $cos(2n\pi+\theta)=cos\theta$ and $sin(2n\pi+\theta)=sin\theta$ (where n=any integer), it follows that,

$Ampz=2n\pi+ampz$ where $-\pi < ampz \leq \pi$

Geometrical representation of modulus and amplitude:

Let us assume that the point $P(x,y)$ represents the complex number $z=x+iy$ . We draw perpendicular $\overline{PM}$ on $\overrightarrow{OX}$ and join $\overline{OP}$ . If $\overline{OP}=r$ and $\angle{XOP}=\theta$ , then from the right-angled triangle $POM$ we get,

$x=r\cos\theta, y=r\sin\theta$

$\tan\theta=\dfrac{\overline{PM}}{\overline{OM}}=\dfrac{y}{x} \Rightarrow \theta=\tan^{-1}\dfrac{y}{x}$

and $r^2=OP^2=OM^2+PM^2=x^2+y^2 \Rightarrow r=\overline{OP}=\sqrt{x^2+y^2}$

$\therefore$ $z+x+iy=r\cos\theta+ir\sin\theta=r(\cos\theta+\sin\theta)$

where, $r=\sqrt{x^2+y^2}=\lvert z\rvert\$

and $\theta=\tan^{-1}\dfrac{y}{x}=Argz$

This form of representation, $z=r(\cos\theta+i\sin\theta)$ where, $r=\lvert z\rvert\$ and $\theta=Argz$ of the complex number $z$ , is called the polar form or the modulus-amplitude form of z.

If the point $P(x,y)$ represents the complex number $z=x+iy$ and $argz=\theta$ then:

a) $0< \theta < \dfrac{\pi}{2}$ when P lies in the first quadrant

b) $\dfrac{\pi}{2}< \theta < \pi$ when P lies in the second quadrant

c) $-\pi < \theta < -\dfrac{\pi}{2}$ when P lies in the third quadrant

d) $-\dfrac{\pi}{2} < \theta < 0$ when P lies in the fourth quadrant

Note:

1) The horizontal axis or the $X$ -axis is also called the real axis. The vertical axis or the $Y$ -axis is called the imaginary axis.

In particular,

when P lies on the positive real axis, principal value of $\theta$ is 0.

when P lies on the negative real axis, principal value of $\theta$ is $\pi$ .

when P lies on the positive imaginary axis, principal value of $\theta$ is $\dfrac{\pi}{2}$ .

when P lies on the negative imaginary axis, principal value of $\theta$ is $-\dfrac{\pi}{2}$ .

2) The argument of $z=0$ i.e., $z=0+i.$ i.e., the origin O, is not defined.

3) It must be kept in mind that $argz$ is the principal value of $\theta$ while $Argz$ is just one of the many values of $\theta$ which satisfy the equations,

$x=\lvert z \rvert \cos\theta$ …(1) and

$y=\lvert z \rvert \sin\theta$ …(2)

Example: Find the amplitude of $\dfrac{i}{1-i}$

Solution: $\dfrac{i}{1-i}=\dfrac{i(1+i)}{(1-i)(1+i)}=\dfrac{i+i^2}{1-i^2}=\dfrac{i-1}{2}=-\dfrac{1}{2}+i.\dfrac{1}{2}$

Clearly, in the z-plane, the point $z=-\dfrac{1}{2}+i.\dfrac{1}{2}=(-\dfrac{1}{2},\dfrac{1}{2})$ lies in the second quadrant.

Hence, if $ampz=\theta$ then, $\tan\theta= \dfrac{\dfrac{1}{2}}{-\dfrac{1}{2}}=-1$ where, $\dfrac{\pi}{2} < \theta \leq \pi$

$\therefore$ $\tan\theta=-1 =\tan(\pi-\dfrac{\pi}{4})=\tan\dfrac{3\pi}{4}$

$\therefore$ $\theta=\dfrac{3\pi}{4}$

$\therefore$ the reqiured amplitude of $\dfrac{i}{1-i}=\dfrac{3\pi}{4}$

Example: Find the amplitude of $\sqrt{12} +6(\dfrac{1-i}{1+i})$

Solution: $\dfrac{1-i}{1+i}=\dfrac{(1-i)^2}{(1+i)(1-i)}=\dfrac{1+i^2-2i}{1-i^2}=\dfrac{1+(-1)-2i}{1-(-1)}=-\dfrac{2i}{2}=-i$

$\therefore$ $z=\sqrt{12} +6(\dfrac{1-i}{1+i})=2\sqrt{3} -6i$

Clearly, in the z-plane, the point $z=2\sqrt{3} -6i=(2\sqrt{3}, -6)$ lies in the fourth quadrant.

Hence, if $ampz=\theta$ then,

$\tan\theta=-\dfrac{6}{2\sqrt{3}}=-\sqrt{3}$ where, $-\dfrac{\pi}{2} < \theta < 0$

$\therefore$ $\tan\theta=-\sqrt{3}=-\tan\dfrac{\pi}{3}=\tan(-\dfrac{\pi}{3})$

$\therefore$ $\theta=-\dfrac{\pi}{3}$ , i.e., the required amplitude of z is $(-\dfrac{\pi}{3})$

Algebra of Complex Numbers:

Let $z_{1}=a+ib$ and $z_{2}=c+id$

Now, in the earlier chapter “Complex Numbers” we have shown that:

For addition: $z_{1} +z_{2}=(a+c)+i(b+d)=A+iB$ where:

$A=a+c$ and $B=b+d$ and both are real.

For subtraction: $z_{1} -z_{2}=a+ib -(-c-id)=(a-c)+i(b-d)=A+iB$ where:

A=a-c $ and $B=b-d$ and both are real.

For multiplication: $z_{1}z_{2}=(a+ib)(c+id)=ac+iad+ibc+i^2bd=(ac-bd) +i(bc+ad)=A+iB$ where:

$A=ac-bd$ and $B=bc+ad$ and both are real.

Note: Proceeding similarly, the product of more than two complex numbers can be expressed in the form $A+iB$

For division: $\dfrac{z_{1}}{z_{2}}=\dfrac{a+ib}{c+id}=\dfrac{(ac+bd)-i(ad-bc)}{c^2+d^2}=\dfrac{ac+bd}{c^2+d^2}+\dfrac{bc-ad}{c^2+d^2}=A+iB$ where:

$A=\dfrac{ac+bd}{c^2+d^2}$ and $B=\dfrac{bc-ad}{c^2+d^2}$ and both are real.

So, it is very clearly that when the four fundamental mathematical operations, viz., addition, subtraction, multiplication and division carried out between two complex numbers, the result is also a complex number of the form $A+iB$ where both $A$ and $B$ are real.

Now, we consider two more algebraic operations:

1) Any integral power of a complex number is a complex number:

Let $z=x+iy$ be a complex number where $x, y$ are real and let $n$ be any integer.

Case I: If $n$ is a positive integer then, $z^n=z.z.z....upto \: n \: factors=(x+iy)(x+iy)...upto \: n \: factors=A+iB$

[ $\because$ the product of more than two complex numbers can be expressed in the form $A+iB$ ] where, $A, B$ are real.

Case II: If $n$ is a negative integer w eassume $n=-m$ where m is taken to be any positive integer.

Then, $z^n=z^{-m}=\dfrac{1}{z^m}=\dfrac{1}{C+iD}$ where $C, D$ are real.

$=\dfrac{C-iD}{(C+iD)(C-iD)}=\dfrac{C}{C^2+D^2}+i(-\dfrac{D}{C^2+D^2})=A+iB$

where, $A=\dfrac{C}{C^2+D^2}$ and $B=-\dfrac{D}{C^2+D^2}$

$\therefore$ any integral power of a complex number is a complex number.

2) Any root of a complex number is a complex number:

Let $z=x+iy$ be a complex number where $x \neq 0, y \neq 0$ are real and let $n$ be any positive integer.

If the $nth$ root of z be $a$ then,

$\sqrt[n]{z}=a \: or, \: \sqrt[n]{x+iy}=a \\ or, \: x+iy=a^n...(1)$

By hypothesis, $x \neq 0, y \neq 0$ Hence, $a^n$ is an imaginary quantity.

Now, $a^n$ can be an imaginary quantity if and only if $a$ is an imaginary number and equation (1) will be satisfies if and only if $a$ is an imaginary number of the form $A+iB$ where $A\neq 0, B \neq 0$ are real.

$\therefore$ any root of a complex number is a complex number.

Properties of Complex Numbers:

1) If $x, y$ are real and $x+iy=o$ then $x=0, y=0$

2) If $x, y, p, q$ are real and $x+iy=p+iq$ then $x=p$ and $y=q$

3) The set of complex numbers satisfies commutative, associative and distributive laws, i.e., if $z_{1}, z_{2} \: and \: z_{3}$ be three complex numbers then,

i) $z_{1}+z_{2}=z_{2}+z_{1}$ [commutative law for addition] and

$z_{1}.z_{2}=z_{2}.z_{1}$ [commutative law for multiplication].

ii) $(z_{1}+z_{2})+z_{3}=z_{1}+(z_{2}+z_{3})$ [associative law for addition] and

$(z_{1}z_{2})z_{3}=z_{1}(z_{2}z_{3})$ [associative law for multiplication].

iii) $z_{1}(z_2{}+z_{3})=z_{1}z_{2}+z_{1}z_{3}$ [distributive law]

4) The sum and product two conjugate complex numbers are both real.

Proof: Let $z=x+iy$ be a complex number where $x, y$ are real.

Then, conjugate of $z=\overline{z}=x-iy$

Now, $z+\overline{z}=x+iy+x-iy=2x$ which is real.

Again, $z.\overline{z}=(x+iy)(x-iy)=x^2-y^2$ which is real.

Note: If $z=x+iy$ then, $\lvert z\rvert\ =\sqrt{x^2+y^2}$

Again, $z.\overline{z}=x^2+y^2$

$\therefore$ $\sqrt{z.\overline{z}}=\sqrt{x^2+y^2}$

Hence, $\lvert z\rvert\ =\sqrt{z.\overline{z}}$

$\Rightarrow$ modulus of a complex quantity =the positive square root of the product of the complex quantity and its conjugate.

5) If the sum and product of two complex numbers are both real, then the complex numbers are conjugate to each other.

Proof: Let $z_{1}=a+ib$ and $z_{2}=c+id$ be two complex numbers where $a, b, c, d$ are al real and $b \neq 0, d \neq 0$

It is given that the sum of $z_{1}$ and $z_{2}$ is real.

$\therefore$ $z_{1}+z_{2}=(a+c)+i(b+d)$ is real.

$\Rightarrow b+d=0 \Rightarrow d=-b$

Also, $z_{1}z_{2}=(ac-bd)+i(ad+bc)$ is real.

$\Rightarrow ad+bc=0 \Rightarrow -ab+bc=0$ [ $\because$ $d=-b$ ]

$\Rightarrow b(c-a)=0 \Rightarrow c=a$ [ $\because$ $b \neq 0$ ]

$\therefore$ $z_{2}=c+id=a+i(-b)=a-ib=\overline{z_{1}}$ which shows that $z_{1}$ and $z_{2}$ are conjugate to each other.

6) For two complex numbers $z_{1}$ and $z_{2}$ ,

$\lvert z_{1}+z_{2}\rvert\ \leq \lvert z_{1}\rvert\ +\lvert z_{2}\rvert\$

Proof: Let $z_{1}=r_{1}(\cos\theta_{1}+i\sin\theta_{2})$ and $z_{2}=r_{2}(\cos\theta_{2}+i\sin\theta_{2})$

$\therefore$ $\lvert z_{1}\rvert =r_{1}$ and $\lvert z_{2}\rvert =r_{2}$

Again, $z_{1}+z_{2}=r_{1}\cos\theta_{1}+r_{1}i\sin\theta_{1}+r_{2}\cos\theta_{2}+r_{2}i\sin\theta_{2}$

$=(r_{1}\cos\theta_{1}+r_{2}\cos\theta_{2})+i(r_{1}\sin\theta_{1}+r_{2}\sin\theta_{2})$

$\therefore$ $\lvert z_{1}+z_{2}\rvert\ =\sqrt{[(r_{1}\cos\theta_{1}+r_{2}\cos\theta_{2})^2+i(r_{1}\sin\theta_{1}+r_{2}\sin\theta_{2})^2]}$

$=\sqrt{[r_{1}^2(\cos^2\theta_{1}+\sin^2\theta_{1})+r_{2}^2(\cos^2\theta_{2}+\sin^2\theta_{2})+2r_{1}r_{2}(\cos\theta_{1}\cos\theta_{2}+\sin\theta_{1}\sin\theta_{2})]}$

$=\sqrt{r_{1}^2+r_{2}^2+2r_{1}r_{2}\cos(\theta_{1}-\theta_{2})}$

Now, $\lvert \cos(\theta_{1}-\theta_{2})\rvert\ \leq 1$

$\therefore$ $\lvert z_{1}+z_{2}\rvert\ \leq \sqrt{r_{1}^2+r_{2}^2+2r_{1}r_{2}}$

$\Rightarrow$ $\lvert z_{1}+z_{2}\rvert\ \leq r_{1}+r_{2}$

$\Rightarrow$ $\lvert z_{1}+z_{2}\rvert\ \leq \lvert z_{1}\rvert\ +\lvert z_{2}\rvert\$

Note:

1) For three complex numbers $z_{1}, z_{2}, z_{3}$ we have,

Proof:

$\lvert z_{1}+z_{2}+z_{3}\rvert\ = \lvert (z_{1}+z_{2})+z_{3}\rvert\ \leq \lvert z_{1}+z_{2}\rvert\ +\lvert z_{3}\rvert\$

$\Rightarrow$ $\lvert z_{1}+z_{2}+z_{3}\rvert\ \leq \lvert z_{1}\rvert\ +\lvert z_{2}\rvert\ +\lvert z_{3}\rvert\$

In general, for n complex numbers $z_{1}, z_{2},...,z_{n}$ we have,

$\lvert z_{1}+z_{2}+...+z_{n}\rvert\ \leq \lvert z_{1}\rvert\ +\lvert z_{2}\rvert\ +... +\lvert z_{n}\rvert\$

2) For two complex numbers $z_{1}$ and $z_{2}$ , we have,

i) $\lvert z_{1}-z_{2}\rvert\ \leq \lvert z_{1}\rvert\ +\lvert z_{2}\rvert\$

ii) $\lvert z_{1}-z_{2}\rvert\ \geq \lvert \lvert z_{1}\rvert\ -\lvert z_{2}\rvert\ \rvert\$

Proof:

i) $\lvert z_{1}-z_{2}\rvert\ =\lvert z_{1}+(-z_{2})\rvert\ \leq \lvert z_{1}\rvert\ +\lvert -z_{2}\rvert\$

$\Rightarrow$ $\lvert z_{1}-z_{2}\rvert\ \leq \lvert z_{1}\rvert\ +\lvert z_{2}\rvert\ [Proved]$

ii) $\lvert z_{1}\rvert\ =\lvert (z_{1}-z_{2})+z_{2}\rvert\ \leq \lvert z_{1}-z_{2}\rvert\ +\lvert z_{2}\rvert\$

$\Rightarrow$ $\lvert z_{1}-z_{2}\rvert\ \geq \lvert z_{1}\rvert\ -\lvert z_{2}\rvert\ ...(1)$

Similarly, $\lvert z_{1} -z_{2}\rvert\ \geq \lvert z_{2}\rvert\ -\lvert z_{1}\rvert\ ...(2)$

From (1) and (2) we get,

$\lvert z_{1} -z_{2}\rvert\ \geq \lvert \lvert z_{1}\rvert\ -\lvert z_{2}\rvert\ \rvert\ [Proved]$

Exercise:

1) Express in the form $A+iB$ where $A, B$ are real:

$a) \: (\dfrac{1+i}{1-i})^3$

$b) \: \dfrac{i}{2+i} +\dfrac{3}{1+4i}$

2) Find the amplitude of the following complex numbers:

$a) \: \dfrac{\sqrt{3}+i}{-1-i\sqrt{3}}$

$b) \: (1+i)(\sqrt{3}+i)$

3) If $z_{1}=-3+4i$ and $z_{2}=12-5i$ , show that,

$a) \: \lvert z_{1}+z_{2}\rvert\ < \lvert z_{1}\rvert\ +\lvert z_{2}\rvert\$

$b) \: \lvert z_{1}z_{2}\rvert\ =\lvert z_{1}\rvert\ \lvert z_{2}\rvert\$

$c) \: \lvert \dfrac{z_{1}}{z_{2}}\rvert =\dfrac{\lvert z_{1}\rvert}{\lvert z_{2}\rvert}$

4) Express $(\sqrt{3}-i)$ in modulus-amplitude form.

5) If $\lvert z_{1}+z_{2}\rvert^2 +\lvert z_{1}-z_{2}\rvert^2 =2[\lvert z_{1}\rvert^2 +\lvert z_{2}\rvert^2]$