1.1 Definition of complex number and Geometrical representation.

1.1 Definition of Complex Number:

A complex number is a number that can be expressed in the form $z=a+bi$, where:

• $a$ (real part) and $b$ (imaginary part) are real numbers.
• $i$ is the imaginary unit, defined as i2=−1i^2 = -1i2=−1.

The number $a+bi$ is composed of:

• The real part, denoted as $\text{Re}(z)=a$.
• The imaginary part, denoted as $\text{Im}(z)=b$.

Complex numbers extend the concept of one-dimensional real numbers to the two-dimensional complex plane.

Examples of Complex Numbers:

• $z=3+4i$ (where the real part is 3 and the imaginary part is 4).
• $z=7-2i$.
• $z=-5+0i$ (purely real number).
• $z=0+6i$ (purely imaginary number).

Geometrical Representation of Complex Numbers:

A complex number can be represented geometrically on a two-dimensional plane known as the complex plane or Argand plane. In this representation:

• The horizontal axis is the real axis (denoted as the x-axis).
• The vertical axis is the imaginary axis (denoted as the y-axis).

For a complex number $z=a+bi$:

• The value $a$ (real part) is plotted on the real axis.
• The value $b$ (imaginary part) is plotted on the imaginary axis.

Thus, the complex number $z=a+bi$ corresponds to the point $(a,b)$ in the complex plane.

Modulus and Argument:

• Modulus (or magnitude) of a complex number $z=a+bi$ is the distance of the point $(a,b)$ from the origin in the complex plane. It is denoted as $|z|$ and calculated as: ∣z∣=a2+b2|z| = \sqrt{a^2 + b^2}∣z∣=a2+b2​
• Argument (or angle) is the angle $\theta$ formed by the line connecting the point $(a,b)$ to the origin with the positive real axis. It is denoted as $\arg (z)$ and can be calculated using: θ=tan⁡−1(ba)\theta = \tan^{-1}\left(\frac{b}{a}\right)θ=tan−1(ab​) (Taking into account the quadrant where the point lies).

Polar Form of Complex Numbers:

A complex number $z=a+bi$ can also be expressed in polar form as:

$$z=r(\cos \theta +i\sin \theta )$$

Where:

• $r=|z|$ is the modulus (or magnitude).
• $\theta =\arg (z)$ is the argument (or angle).

This can also be written in the more compact Euler’s form:

z=reiθz = r e^{i \theta}z=reiθ

Where eiθ=cos⁡θ+isin⁡θe^{i \theta} = \cos \theta + i \sin \thetaeiθ=cosθ+isinθ.

Geometrical Representation Example:

For the complex number $z=3+4i$:

• Real part $a=3$ and imaginary part $b=4$.
• It is represented as the point $(3,4)$ in the complex plane.
• The modulus is: ∣z∣=32+42=9+16=5|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5∣z∣=32+42​=9+16​=5
• The argument is: θ=tan⁡−1(43)≈53.13∘\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circθ=tan−1(34​)≈53.13∘

This geometric interpretation allows us to visualize complex numbers as vectors in the complex plane, with the modulus representing the length of the vector and the argument representing the angle of rotation from the real axis.

1.2 Conjugate and absolute value of complex number

1.2 Conjugate and Absolute Value of a Complex Number

1. Conjugate of a Complex Number:

The conjugate of a complex number $z=a+bi$ (where $a$ is the real part and $b$ is the imaginary part) is denoted as z‾\overline{z}z, and it is obtained by changing the sign of the imaginary part.

So, the conjugate of $z=a+bi$ is:

z‾=a−bi\overline{z} = a – biz=a−bi

Where:

• The real part $a$ remains the same.
• The imaginary part $bi$ becomes $-bi$.

Examples:

• If $z=3+4i$, then the conjugate z‾=3−4i\overline{z} = 3 – 4iz=3−4i.
• If $z=5-2i$, then the conjugate z‾=5+2i\overline{z} = 5 + 2iz=5+2i.
• If $z=-7+i$, then the conjugate z‾=−7−i\overline{z} = -7 – iz=−7−i.

Key Properties of the Conjugate:

• z+z‾=2az + \overline{z} = 2az+z=2a (the real part of the complex number is doubled).
• z⋅z‾=a2+b2z \cdot \overline{z} = a^2 + b^2z⋅z=a2+b2, which is the square of the modulus (absolute value) of the complex number.

2. Absolute Value (or Modulus) of a Complex Number:

The absolute value or modulus of a complex number $z=a+bi$ is a measure of the distance from the origin to the point $(a,b)$ in the complex plane. It is denoted by $|z|$ and is defined as:

∣z∣=a2+b2|z| = \sqrt{a^2 + b^2}∣z∣=a2+b2​

Where:

• $a$ is the real part.
• $b$ is the imaginary part.

The modulus is always a non-negative real number.

Examples:

• For $z=3+4i$, the modulus is: ∣z∣=32+42=9+16=25=5|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5∣z∣=32+42​=9+16​=25​=5
• For $z=-1+i$, the modulus is: ∣z∣=(−1)2+12=1+1=2|z| = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}∣z∣=(−1)2+12​=1+1​=2​
• For $z=5-12i$, the modulus is: ∣z∣=52+(−12)2=25+144=169=13|z| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13∣z∣=52+(−12)2​=25+144​=169​=13

Relation between Conjugate and Modulus:

The product of a complex number and its conjugate gives the square of the modulus:

z⋅z‾=(a+bi)(a−bi)=a2−(bi)2=a2+b2=∣z∣2z \cdot \overline{z} = (a + bi)(a – bi) = a^2 – (bi)^2 = a^2 + b^2 = |z|^2z⋅z=(a+bi)(a−bi)=a2−(bi)2=a2+b2=∣z∣2

Thus:

∣z∣=z⋅z‾|z| = \sqrt{z \cdot \overline{z}}∣z∣=z⋅z​

Geometric Interpretation:

• The conjugate of a complex number is its reflection across the real axis in the complex plane.
• The modulus represents the distance from the origin $(0,0)$ to the point $(a,b)$, which corresponds to the complex number $z$.

1.3 Square roots of complex number

1.3 Square Roots of a Complex Number

To find the square root of a complex number $z=a+bi$, where $a$ is the real part and $b$ is the imaginary part, you can use the following method, which involves expressing the complex number in polar form and applying some algebra.

Step-by-Step Method for Finding Square Roots:

1. Express the complex number in polar form: A complex number $z=a+bi$ can be written in polar form as:

   $$z=r\left(\cos \theta +i\sin \theta \right)$$Where:

   • r=∣z∣=a2+b2r = |z| = \sqrt{a^2 + b^2}r=∣z∣=a2+b2​ is the modulus (or absolute value) of the complex number.
   • θ=arg⁡(z)=tan⁡−1(ba)\theta = \arg(z) = \tan^{-1} \left( \frac{b}{a} \right)θ=arg(z)=tan−1(ab​) is the argument (or angle) of the complex number.
2. Take the square root of the modulus: The square root of the modulus $r$ is:

   r\sqrt{r}r​

3. Divide the argument by 2: Since you’re looking for the square root, divide the argument $\theta$ by 2:

   θ2\frac{\theta}{2}2θ​

4. Use the square root formula for complex numbers: The square roots of $z=r\left(\cos \theta +i\sin \theta \right)$ are given by:

   ±r(cos⁡θ2+isin⁡θ2)\pm \sqrt{r} \left( \cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right)±r​(cos2θ​+isin2θ​)There are two square roots of a complex number, corresponding to the $\pm$ signs.

Example 1: Finding the square root of a complex number $z=3+4i$

1. Find the modulus $r$:

   r=32+42=9+16=25=5r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5r=32+42​=9+16​=25​=5

2. Find the argument $\theta$:

   θ=tan⁡−1(43)≈53.13∘\theta = \tan^{-1} \left( \frac{4}{3} \right) \approx 53.13^\circθ=tan−1(34​)≈53.13∘(This can also be expressed in radians as $\theta \approx 0.927$ radians).

3. Take the square root of the modulus:

   r=5≈2.236\sqrt{r} = \sqrt{5} \approx 2.236r​=5​≈2.236

4. Divide the argument by 2:

   θ2≈53.13∘2≈26.57∘\frac{\theta}{2} \approx \frac{53.13^\circ}{2} \approx 26.57^\circ2θ​≈253.13∘​≈26.57∘

5. Apply the square root formula: The two square roots are:

   ±5(cos⁡26.57∘+isin⁡26.57∘)\pm \sqrt{5} \left( \cos 26.57^\circ + i \sin 26.57^\circ \right)±5​(cos26.57∘+isin26.57∘)Converting back to rectangular form, this gives:

   $$\pm \left(2+i\right)$$So the square roots of $3+4i$ are approximately:

   z1=2+iandz2=−2−iz_1 = 2 + i \quad \text{and} \quad z_2 = -2 – iz1​=2+iandz2​=−2−i

Example 2: Finding the square root of a purely imaginary number $z=0+6i$

1. Find the modulus $r$:

   r=02+62=36=6r = \sqrt{0^2 + 6^2} = \sqrt{36} = 6r=02+62​=36​=6

2. Find the argument $\theta$: For $z=0+6i$, the argument is 90∘90^\circ90∘ (or π2\frac{\pi}{2}2π​ radians).
3. Take the square root of the modulus:

   r=6≈2.449\sqrt{r} = \sqrt{6} \approx 2.449r​=6​≈2.449

4. Divide the argument by 2:

   θ2=90∘2=45∘\frac{\theta}{2} = \frac{90^\circ}{2} = 45^\circ2θ​=290∘​=45∘

5. Apply the square root formula: The square roots are:

   ±6(cos⁡45∘+isin⁡45∘)\pm \sqrt{6} \left( \cos 45^\circ + i \sin 45^\circ \right)±6​(cos45∘+isin45∘)Converting back to rectangular form, this gives:

   $$\pm \left(1.732+1.732i\right)$$So the square roots of $0+6i$ are approximately:

   z1=1.732+1.732iandz2=−1.732−1.732iz_1 = 1.732 + 1.732i \quad \text{and} \quad z_2 = -1.732 – 1.732iz1​=1.732+1.732iandz2​=−1.732−1.732i

1.4 Polar and Exponential forms of complex numbers

Polar Form of Complex Numbers

A complex number $z=a+bi$ (where $a$ is the real part and $b$ is the imaginary part) can be represented in polar form as:

$$z=r(\cos \theta +i\sin \theta )$$

Where:

• r=∣z∣=a2+b2r = |z| = \sqrt{a^2 + b^2}r=∣z∣=a2+b2​ is the magnitude (modulus) of the complex number.
• $\theta$ is the argument (angle) of the complex number, measured in radians. It can be calculated using θ=tan⁡−1(ba)\theta = \tan^{-1} \left( \frac{b}{a} \right)θ=tan−1(ab​).

Exponential Form of Complex Numbers

The exponential form of a complex number is derived from Euler’s formula:

eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos \theta + i \sin \thetaeiθ=cosθ+isinθ

Thus, the polar form $z=r(\cos \theta +i\sin \theta )$ can be written in exponential form as:

z=reiθz = re^{i\theta}z=reiθ

Where:

• $r$ is the magnitude (modulus) of the complex number.
• $\theta$ is the argument (angle) of the complex number.

1.5 nth roots of complex numbers using DeMoivre’s Theorem

$n$-th Roots of Complex Numbers Using De Moivre’s Theorem

De Moivre’s Theorem provides a convenient way to find the powers and roots of complex numbers when they are in polar or exponential form.

De Moivre’s Theorem

For any complex number $z=r(\cos \theta +i\sin \theta )$, De Moivre’s Theorem states:

zn=rn(cos⁡(nθ)+isin⁡(nθ))z^n = r^n \left( \cos(n\theta) + i \sin(n\theta) \right)zn=rn(cos(nθ)+isin(nθ))

This theorem is useful for calculating powers of complex numbers. To find the $n$-th roots of a complex number, we use an adaptation of this theorem.

$n$-th Roots of a Complex Number

Given a complex number $z=r(\cos \theta +i\sin \theta )$, the $n$-th roots of $z$ are given by:

z1n=r1n(cos⁡θ+2kπn+isin⁡θ+2kπn)for k=0,1,2,…,n−1z^{\frac{1}{n}} = r^{\frac{1}{n}} \left( \cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n} \right) \quad \text{for} \, k = 0, 1, 2, \dots, n-1zn1​=rn1​(cosnθ+2kπ​+isinnθ+2kπ​)fork=0,1,2,…,n−1

Where:

• $r$ is the modulus (magnitude) of the complex number.
• $\theta$ is the argument (angle) of the complex number.
• $n$ is the degree of the root.
• $k$ represents different solutions, which vary by multiples of 2πn\frac{2\pi}{n}n2π​. There are $n$ distinct roots corresponding to $k=0,1,2,\dots ,n-1$.

Steps to Find the $n$-th Roots

1. Convert the complex number $z=a+bi$ into polar form:

   $$z=r(\cos \theta +i\sin \theta )$$where r=∣z∣=a2+b2r = |z| = \sqrt{a^2 + b^2}r=∣z∣=a2+b2​ and θ=tan⁡−1(ba)\theta = \tan^{-1} \left( \frac{b}{a} \right)θ=tan−1(ab​).

2. Apply the $n$-th root formula:

   z1n=r1n(cos⁡θ+2kπn+isin⁡θ+2kπn)z^{\frac{1}{n}} = r^{\frac{1}{n}} \left( \cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n} \right)zn1​=rn1​(cosnθ+2kπ​+isinnθ+2kπ​)

3. Compute the distinct roots by varying $k$ from 0 to $n-1$. Each value of $k$ gives a different root.

Example

Find the cube roots of $z=8$ (which can be written as $z=8+0i$).

1. The modulus $r=|8+0i|=8$ and the argument θ=tan⁡−1(0/8)=0\theta = \tan^{-1}(0/8) = 0θ=tan−1(0/8)=0.
2. The cube roots are:

   z13=813(cos⁡0+2kπ3+isin⁡0+2kπ3)z^{\frac{1}{3}} = 8^{\frac{1}{3}} \left( \cos \frac{0 + 2k\pi}{3} + i \sin \frac{0 + 2k\pi}{3} \right)z31​=831​(cos30+2kπ​+isin30+2kπ​)Simplifying:

   z13=2(cos⁡2kπ3+isin⁡2kπ3)z^{\frac{1}{3}} = 2 \left( \cos \frac{2k\pi}{3} + i \sin \frac{2k\pi}{3} \right)z31​=2(cos32kπ​+isin32kπ​)

3. For $k=0$:

   z0=2(cos⁡0+isin⁡0)=2(1+0i)=2z_0 = 2 (\cos 0 + i \sin 0) = 2 (1 + 0i) = 2z0​=2(cos0+isin0)=2(1+0i)=2For $k=1$:

   z1=2(cos⁡2π3+isin⁡2π3)=2(−12+i32)=−1+i3z_1 = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) = 2 \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = -1 + i \sqrt{3}z1​=2(cos32π​+isin32π​)=2(−21​+i23​​)=−1+i3​For $k=2$:

   z2=2(cos⁡4π3+isin⁡4π3)=2(−12−i32)=−1−i3z_2 = 2 \left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right) = 2 \left( -\frac{1}{2} – i \frac{\sqrt{3}}{2} \right) = -1 – i \sqrt{3}z2​=2(cos34π​+isin34π​)=2(−21​−i23​​)=−1−i3​

Thus, the three cube roots of 8 are $2$, −1+i3-1 + i \sqrt{3}−1+i3​, and −1−i3-1 – i \sqrt{3}−1−i3​.