Complex Numbers - MathsTips.com

There are a few sets of numbers which we are already familiar with. These numbers are called real numbers. The aggregate of the sets of rational and irrational numbers is called the set of real numbers. The basic and most important property of any real number is that its square is always positive or non-negative. Because of this property, it was realized that real numbers are not sufficient for solving all algebraic problems. Hence, there was the need to invent a new set of numbers. These numbers are different from real numbers. This new set of a numbers are called complex numbers or imaginary numbers.

For example, the quadratic equation $ax^2+bx+c=0$ , where $a,b,c \in \mathbb{R}$ , (where $\mathbb{R}$ represents the set of real numbers), cannot be solved if $b^2-4ac<0$ , hence the need for complex numbers.

The equation $x^2+1=0$ cannot be solved in the real number system. Hence Euler introduced the symbol $i$ to represent $\sqrt{(-1)}$ and he said $i^2=(-1)$ .

$i$ is a solution to the equation $x^2+1=0$ .

Thus it has two solutions, $x=\pm i$ where $i=\sqrt{(-1)}$ .

This number $i$ is called an imaginary number or complex number.

Definition of a Complex Number

If an ordered pair $(x,y)$ of two real numbers $x$ and $y$ is represented by the symbol $x+iy$ where $i=\sqrt{(-1)}$ , then the ordered pair is called a complex or imaginary number.

Thus, a complex number $z$ is defined as an ordered pair of real numbers and written as $z=a+ib$ where $a,b \in \mathbb{R}$ and $i=\sqrt{(-1)}$

For example, $(3+5i), (2-3i)$ etc. are all complex numbers.

Real and Imaginary parts of Complex Number

If $z=a+ib$ is a complex number, then the real part of $z$ , is denoted by $Re(z)=a$ and the imaginary part is denoted by $Im(z)=b$

In a complex number $z=a+ib$ when the real part $Re(z)=a$ is zero or when $a=0$ , then the number is said to be purely imaginary.

Similarly, in a complex number, when the imaginary part, i.e., $Im(z)=b$ is zero, or when $b=0$ , then the number is said to be purely real.

For example, $3i=0+i.3$ and here $a=0, b=3$ .

So, $3i$ is a purely imaginary number.

Again, $5=5+i.0$ and here $a=5, b=0$ .

So $5$ is a purely real number.

Negative of Complex Number

If $z=a+ib$ then $(-z)=-(a+ib)$

$\therefore$ $(-z)$ is the negative of the complex number $z$ , also called the additive inverse of z.

Powers of i

We already know that $i=\sqrt{(-1)}$ . Similarly we can write the higher powers of $i$ as follows:

$i^{4} =(i^2)^2=(-1)^2=1$

$i^{5}=i ^{4} \times i=i$

$i^{6}= i^{4} \times i^{2}=1 \times (-1) =(-1)$

$i^8=(i^{4})^{2}=1$

$i^9=i^8 \times i=i$

This shows that the values repeat themselves in cycles of four.

Complex Number Operations

1. Addition:

Let there be two complex numbers, $z_{1}=a+ib$ and $z_{2}=c+id$ We add the real parts and imaginary parts separately.

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

For example: Say, $z_{1}=2+3i$ and $z_{2}=5+6i$

$\therefore$ $z_{1} +z_{2}=(2+5)+i(3+6)$

$\therefore$ $z=z_{1} +z_{2}=7+9i$

2. Subtraction:

We change one complex number into its additive inverse and then add the two numbers. Let there be two complex numbers, $z_{1}=a+ib$ and $z_{2}=c+id$

$\therefore$ $z_{1} -z_{2}= z_{1}+ (-z_{2})=a+ib -(-c-id)=(a-c)+i(b-d)$

$\therefore$ $z=z_{1} -z_{2}=(a-c)+i(b-d)$

For example: Say, $z_{1}=6+7i$ and $z_{2}=3+2i$

$\therefore$ $z_{1} -z_{2}=z_{1}+(-z_{2})=6+7i -(-3-2i)=(6-3)+i(7-2)=3+5i$

$\therefore$ $z=z_{1} -z_{2}=3+5i$

3. Multiplication:

Let there be two complex numbers, $z_{1}=a+ib$ and $z_{2}=c+id$

$\therefore$ $z_{1}z_{2}=(a+ib)(c+id)$

$z_{1}z_{2}=ac+aid+cib+i^{2}bd \\ \vspace{5mm} \\ =ac+i(ad+cb)-bd \\ \vspace{5mm} \\ =(ac-bd) +i(ad+cb)$

For example: Say, $z_{1}=2+3i$ and $z_{2}=5+4i$

$\therefore$ $z_{1}z_{2}=(2+3i)(5+4i)$

$=10+8i+15i+12i^2 \\ \vspace{5mm} \\ =10-12+i(8+15) \\ \vspace{5mm} \\ =(-2+23i)$

4. Division:

Let there be two complex numbers, $z_{1}=a+ib$ and $z_{2}=c+id$ . Division of these two complex numbers is defined as:

$\dfrac{z_{1}}{z_{2}} =\dfrac{a+ib}{c+id}$

We rationalise the denominator, $\dfrac{z_{1}}{z_{2}} =\dfrac{(a+ib) \times (c-id)}{(c+id) \times (c-id)}$

$=\dfrac{(ac-aid+cib-i^{2}db)}{c^2-i^{2}d^{2}} \\ \vspace{5mm} \\ =\dfrac{(ac+bd)-i(ad-bc)}{c^2+d^2}$

For example: Say, $z_{1}=2+3i$ and $z_{2}=3+4i$

$\dfrac{z_{1}}{z_{2}}=\dfrac{2+3i}{3+4i} \\ \vspace{5mm} \\ =\dfrac{(2+3i) \times (3-4i)}{(3+4i) \times (3-4i)} \\ \vspace{5mm} \\ =\dfrac{6-8i+9i-12i^2}{9-16i^2} \\ \vspace{5mm} \\ =\dfrac{(6+12)+i(9-8)}{(9+16)} \\ \vspace {5mm} \\ =\dfrac{13+i}{25}$ $\therefore$ $\dfrac{z_{1}}{z_{2}}=\dfrac{13}{25}+\dfrac{i}{25}$

5. Conjugate of Complex Number:

When two complex numbers only differ in the sign of their complex parts, they are said to be the conjugate of each other. The conjugate is denoted as $\bar{z}$ .

If $z=a+ib,$ then $\bar{z}=a-ib$

For example, if $z=3+4i$ , the conjugate of $z$ is $\bar{z}=3-4i$

Properties of Conjugates:

$\overline{(\overline{z})}=z$ , i.e., conjugate of conjugate gives the original complex number

$\overline{(z_{1}+z_{2})}=\bar{z_{1}}+\bar{z_{2}}$

$\overline{(z_{1}-z_{2})}=\bar{z_{1}}+\bar{z_{2}}$

$\overline{(z_{1}z_{2})}=\bar{z_{1}} \times \bar{z_{2}}$

$\overline{(\dfrac{z_{1}}{z_{2}})}=\dfrac{\bar{z_{1}}}{\bar{z_{2}}}$

6. Modulus of a Complex Number:

The absolute value or modulus of a complex number, $z=a+ib$ is denoted by $\lvert z\rvert\$ and is defined as:

$\lvert z\rvert\ =\sqrt{a^2+b^2}=\sqrt{{Re(z)}^2+{Im(z)}^2}$

Here, $a=Re(z), b=Im(z)$

For example: If $z=2+3i$ $\lvert z\rvert\ =\sqrt{2^2+3^2}=\sqrt{4+9}=\sqrt{13}$

$\therefore$ $\lvert z\rvert\ =\sqrt{13}$

Properties of Modulus:

$\lvert -z\rvert\ = \lvert z\rvert\$

$\lvert z\rvert\ =0$ only if $z=0$ when $Re(z)=Im(z)=0$

$\lvert z_{1}z_{2}\rvert\ = \lvert z_{1}\rvert\$ $\lvert z_{2}\rvert\$

$\lvert \dfrac{z_{1}}{z_{2}}\rvert\ =\lvert z_{1}\rvert\ \div \lvert z_{2}\rvert\$

$\lvert \overline{z}\rvert\ = \lvert z\rvert\$

7. Equality of Complex Numbers:

Two complex numbers $z_{1}=a+ib, z_{2}=a-ib$ are said to be equal if and only if $a=c$ and $b=d$

Illustrations:

1. If $x+iy=\dfrac{a+ib}{a-ib}$ , then prove that $x^2+y^2=1$

Solution:

$x+iy=\dfrac{a+ib}{a-ib}$

$\therefore$ $\lvert x+iy\rvert\ =\dfrac{\lvert a+ib\rvert\ }{\lvert a-ib\rvert\ }$

$\Rightarrow$ $\sqrt{x^2+y^2}=\dfrac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}$

$\Rightarrow$ $\sqrt{x^2+y^2}=1$

$\Rightarrow$ $x^2+y^2=1 (Proved)$

2. Find the real numbers $x$ and $y$ if $(x-iy)(3+5i)$ is the conjugate of $(-6-24i)$

Solution:

$(x-iy)(3+5i)=3x+5xi-3iy-5i^2y=3x+5y+i(5x-3y)$

The conjugate of $(-6-24i)$ is $(-6+24i)$

According to the problem,

$3x+5y+i(5x-3y) =-6+24i$

$\therefore$ $3x+5y=-6...(1)$

$\therefore$ $5x-3y=24...(2)$

Solving equations (1) and (2) we get,

$x=3$ and $y=-3$ (Answer)

3. Find the values of $x$ and $y$ if the complex numbers $z_{1}=6y+5i$ and $z_{2}=12+10xi$ are equal

Solution:

$z_{1}=z_{2}$

$\Rightarrow 6y+5i=12+10xi...(1)$

$\therefore$ $Re(z_{1})=Re(z_{2})$

and $Im(z_{1})=Im(z_{2})$

$\therefore$ from (1) we have,

$6y=12 \Rightarrow y=2$

$10x=5 \Rightarrow x=\dfrac{1}{2}$

Answer: $x=\dfrac{1}{2} \: y=2$

4. Find the modulus of the following:

$\dfrac{3}{5} +\dfrac{4}{5}i$
$\dfrac{\sqrt{7}}{2} -\dfrac{3}{2}i$

Solutions:

4a) $z_{1} =\dfrac{3}{5}+\dfrac{4}{5}i$

$\therefore$ modulus of $z_{1}$ is given by $\lvert z_{1}\rvert\ =\lvert \dfrac{3}{5}+ \dfrac{4}{5}i\rvert\ =\sqrt{(\dfrac{3}{5})^2+(\dfrac{4}{5})^2}=\sqrt{\dfrac{9}{25}+\dfrac{16}{25}}$

$=\sqrt{\dfrac{25}{25}}=1$

4b) $z_{2} =\dfrac{\sqrt{7}}{2}+\dfrac{3}{2}i$

$\therefore$ modulus of $z_{2}$ is given by $\lvert z_{2}\rvert\ =\lvert \dfrac{\sqrt{7}}{2} +\dfrac{3}{2}i\rvert\ =\sqrt{(\dfrac{\sqrt{7}}{2})^2+(\dfrac{-3}{2})^2}=\sqrt{\dfrac{7}{4}+\dfrac{9}{4}}$

$=\sqrt{\dfrac{16}{4}}=\sqrt{4}=2$

5. If $z=x+iy$ and $\lvert z-1\rvert\ + \lvert z+1\rvert\ =4$ , show that, $3x^2+4y^2=12$

Solution:

$\lvert z-1\rvert\ + \lvert z+1\rvert\ =4$ or, $\lvert x+iy-1\rvert\ +\lvert x+iy+1\rvert\ =4$

or, $\lvert (x-1)+iy\rvert\ +\lvert (x+1)+iy\rvert\ =4$

or, $\sqrt{(x-1)^2+y^2} +\sqrt{(x+1)^2+y^2} =4$

or, $\sqrt{(x-1)^2+y^2} = 4- \sqrt{(x+1)^2+y^2}$

or, $(x+1)^2+y^2=16+(x-1)^2+y^2-8\sqrt{(x-1)^2+y^2}$

or, $8\sqrt{(x-1)^2+y^2} =16+(x-1)^2+y^2-(x+1)^2-y^2=16-4x$

[ $\because$ $(a+b)^2-(a-b)^2=4ab$ ]

or, $2\sqrt{(x-1)^2+y^2} =4-x$

or, $4[(x-1)^2+y^2]=(4-x)^2$ [ squaring both sides ]

or, $4(x^2-2x+1+y^2) =16+x^2-8x$

or, $3x^2+4y^2=12$

Exercise:

If $a+ib=\dfrac{(x+i)^2}{2x-i}$ , then prove that $a^2+b^2=\dfrac{(x^2+1)^2}{4x^2+1}$
If $x+iy=\sqrt{\dfrac{1+i}{1-i}}$ , then prove that $x^2+y^2=1$
Express $\dfrac{1+7i}{(2-i)^2}$ in the form $a+ib$ . Also find the conjugate and modulus of it.
Find the modulus of $2\sqrt{2}-\dfrac{13}{\sqrt{7}}i$
Find the absolute value of $(-3+4i)$
Find $x$ and $y$ if $(3x-7)+5iy=2y+3-4(1-x)i$
Simplify $1+i^2+i^4+i^6$
Show that $[i^{37}-(\dfrac{1}{i})^41]^3=-8i$
For any complex number, $z$ , prove that: $Re(z)=\dfrac{z+\overline{z}}{2}$ and $Im(z)=\dfrac{z-\overline{z}}{2i}$
For any two complex numbers $z_{1}$ and $z_{2}$ , prove that: $Re(z_{1}z_{2})=Re(z_{1})Re(z_{2})-Im(z_{1})Im(z_{2})$
Find the real values of $x$ and $y$ for which $\dfrac{x-1}{3+i} +\dfrac{y-1}{3-i}=i$
If $z_{1}=2-i$ and $z_{2}=1+i$ find $\lvert \dfrac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i}\rvert\$

Trackbacks

Quadratic Equations – Maths Tips says:

March 23, 2014 at 10:39 am

[…] which are complex roots. To know more about complex numbers and ‘i’ refer Complex Numbers. […]

More on Complex Numbers – Maths Tips says:

April 18, 2014 at 4:23 am

[…] earlier chapter on Complex Numbers explained what we mean by a complex number together with a few properties of complex numbers. Basic […]

Definition of a Complex Number

Real and Imaginary parts of Complex Number

Negative of Complex Number

Powers of i

Complex Number Operations

1. Addition:

2. Subtraction:

3. Multiplication:

4. Division:

5. Conjugate of Complex Number:

Properties of Conjugates:

6. Modulus of a Complex Number:

Properties of Modulus:

7. Equality of Complex Numbers:

Illustrations:

Exercise:

Trackbacks

Leave a Reply Cancel reply