Euler Form of Complex Number

Introduction to Complex Numbers

A complex number is a number of the form $a + ib$, where $a$ and $b$ are real numbers and $i = \sqrt{-1}$. Complex numbers are denoted by letters like $z$, $z_1$, or $z_2$.

Example: $z = 5 + 2i$
Here, $5$ is the real part, denoted as $\operatorname{Re}(z)$, and $2$ is the imaginary part, denoted as $\operatorname{Im}(z)$.

Euler’s Formula: Definition and Significance

Euler’s formula, introduced by Leonhard Euler, connects complex numbers with exponential and trigonometric functions. It states:

$e^{i\theta} = \cos \theta + i \sin \theta$

This formula simplifies operations on complex numbers such as multiplication, division, and finding powers or roots.

Euler form of a complex number

Euler’s formula establishes a fundamental relationship between trigonometric functions and exponential functions. Geometrically, it bridges two representations of the same unit complex number in the complex plane. This representation is extremely convenient and simplifies many calculations.

The polar form of a complex number is:

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

In Euler form, the $(\cos \theta + i \sin \theta)$ part of the polar form is represented using the exponential function:

$z = r e^{i \theta}$

Here, $r = |z|$ is the modulus and $\theta = \arg(z)$ is the argument of the complex number.

Euler's Formula: Derivation

We know the expansion of $e^x$ is:

$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$

Replacing $x$ by $ix$, we get:

$\begin{aligned} e^{ix} &= 1 + \frac{ix}{1!} + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \ldots \\ &= 1 + \frac{ix}{1!} - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \ldots \end{aligned}$

Rearranging real and imaginary terms:

$e^{ix} = \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots \right) + i \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots \right)$

We notice that the first bracket is the expansion of $\cos x$ and the second bracket is the expansion of $\sin x$, so:

$e^{ix} = \cos x + i \sin x$

Similarly:

$e^{-i\theta} = \cos \theta - i \sin \theta$

Euler forms simplify algebra for complex numbers, especially for multiplication, division, and powers. Any complex number can be expressed in different forms:

• Cartesian form: $z = x + iy$

• Polar form: $z = r(\cos \theta + i \sin \theta)$

• Euler form: $z = |z| e^{i\theta}$

1. Multiplication of Two Complex Numbers

Let $z = |z| e^{i\theta_1} \quad \text{and} \quad w = |w| e^{i\theta_2}$

Multiplying these two complex numbers:

$z \cdot w = |z| e^{i\theta_1} \cdot |w| e^{i\theta_2} = |z| \cdot |w| \, e^{i(\theta_1+\theta_2)}$

This shows that in Euler form, the moduli multiply and the arguments add.

2. Division of Two Complex Numbers

For two complex numbers in Euler form:

$z = |z| e^{i\theta_1}, \quad w = |w| e^{i\theta_2}$

the division is given by:

$\frac{z}{w} = \frac{|z|}{|w|} \, e^{i(\theta_1 - \theta_2)}$

Here, the moduli divide and the arguments subtract, simplifying division significantly.

3. Logarithm of a Complex Number

For a complex number in Euler form:

$z = |z| e^{i\theta}$

the natural logarithm is:

$\begin{aligned} \log_e(z) &= \log_e\left(|z| e^{i\theta}\right) \\ &= \log_e(|z|) + \log_e\left(e^{i\theta}\right) \\ &= \log_e(|z|) + i \arg(z) \end{aligned}$

This property makes handling powers and roots of complex numbers much easier.

4. Euler’s Identity

Euler’s identity is one of the most elegant equations in mathematics:

$e^{i \pi} + 1 = 0 \quad \text{or equivalently} \quad e^{i\pi} = -1$

It beautifully connects five fundamental mathematical constants: $e$, $i$, $\pi$, $1$, and $0$, and demonstrates the deep relationship between exponential and trigonometric functions.

Euler’s identity is derived by substituting $x = \pi$ in Euler’s formula:

$e^{ix} = \cos x + i \sin x$

and simplifying, showing its connection to the unit circle in the complex plane.

Solved Examples Based On the Euler Form of a Complex Number

Example 1: Euler's form of $z=-\sqrt{3}-i$ is

Solution:

Euler's Form of a Complex Number -

$z=r e^{i \theta}$

where r denotes the modulus of z and $\theta$ denotes the argument of z .

Now, $z=-\sqrt{3}-i$

$\therefore r=|z|=\sqrt{3+1}=2$

and Z being in 3rd quadrant, its arg(z) will be :

$-\pi+\tan ^{-1}\left|\frac{-1}{-\sqrt{3}}\right|=-\pi+\frac{\pi}{6}=-\frac{5 \pi}{6}$

$Z=2 e^{-i \frac{5 \pi}{6}}$

Example 2: $\mathrm{z}=\frac{16}{1+i \sqrt{3}}$ , its Euler form is?

Solution:

We simplify z, and for that, we normalize the denominator

$\mathrm{z}=\frac{16}{1+\mathrm{i} \sqrt{3}} \cdot \frac{1-\mathrm{i} \sqrt{3}}{1-\mathrm{i} \sqrt{3}}=4(1-\mathrm{i} \sqrt{3})$

Now we see it lies in the 4th quadrant, so the argument is going to be -ve.

First we find r = |z| = 4·2=8
$\theta=\arg (\mathrm{z})=\tan ^{-1}(-\sqrt{3})=\frac{-\pi}{3}$
So euler form $=r e^{\mathrm{i} \theta}=8 \mathrm{e}^{-\frac{\pi}{3} \mathrm{i}}$

Hence, the answer is $8 e^{-\frac{\pi}{3}}$.

Example 3: Real part of $e^{e^{10}}$ is equal to:

Solution:

$e^{e^{i \theta}} = e^{\cos \theta + i \sin \theta} = e^{\cos \theta} \cdot e^{i \sin \theta}$

$= e^{\cos \theta} \cdot [\cos (\sin \theta) + i \sin (\sin \theta)]$

$= e^{\cos \theta} \cdot \cos (\sin \theta) + i , e^{\cos \theta} \cdot \sin (\sin \theta)$

$\text{(Real part)} \quad \text{(Imaginary part)}$

Example 4: Euler's form of $z=\frac{1-7 i}{(2+i)^2}$ will be

Solution:

Euler's Form of a Complex Number -

$z=r e^{i \theta}$

where $r$ denotes the modulus of $z$ and $\theta$ denotes the argument of $z$.

Now,

$Z=\frac{1-7 i}{(2+i)^2}=\frac{1-7 i}{4-1+4 i}=\frac{1-7 i}{3+4 i} \times \frac{3-4 i}{3-4 i}=\frac{3-28-25 i}{25}$

$\therefore z=-1-i$

$\therefore r=|z|=\sqrt{1+1}=\sqrt{2} \text { and } \arg (\mathrm{z})=\tan ^{-1}\left|\frac{-1}{-1}\right|-\pi$

$r=\sqrt{2} \text { and } \arg (\mathrm{z})=\frac{-3 \pi}{4} \Rightarrow z=\sqrt{2} e^{-i \frac{3 \pi}{4}}$

Example 5: If $z$ and $w$ are two complex numbers such that $|z w|=1$ and $\arg (z)-\arg (w)=\frac{\pi}{2}$ then $\bar{z} w=-i$ equals to:

Solution:

Euler's Form of a Complex Number -

$z=r e^{i \theta}$

- wherein

r denotes the modulus of $z$ and $\theta$ denotes the argument of $z$.

Polar Form of a Complex Number -

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

- wherein

$r=$ modulus of $z$ and $\theta$ is the argument of $z$

Now,

$|z w|=1$ and $\arg (z)-\arg (w)=\frac{\pi}{2}$

Let $|z|=r$ $\Rightarrow z=r e^{i \theta}$

$|\omega|=\frac{1}{r}$ $=>\omega=\frac{1}{r} e^{i \phi}$

$\arg (z)-\arg (w)=\frac{\pi}{2}$

$\theta-\phi=\frac{\pi}{2}$

$\theta=\frac{\pi}{2}+\phi$

$z \bar{\omega}=r e^{i \theta} \cdot \frac{1}{r} e^{-i \phi}$

$=r e^{i(\theta-\phi)}$

$=r e^{i\left(\frac{\pi}{2}+\phi-\phi\right)}$

$=r e^{i\left(\frac{\pi}{2}\right)}$

$=\cos \left(\frac{\pi}{2}\right)+i \sin \left(\frac{\pi}{2}\right)$

$=0+i .1 = i$

List of topics related to the Euler form of a Complex Numbers

Below is a list of important topics related to the Euler form of a complex number, covering different key concepts, formulas of complex numbers for better understanding and practice.

NCERT Resources

Here you can find NCERT resources for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations, including detailed solutions, exemplar problems, and summary notes to strengthen your concepts and enhance practice.

NCERT Solutions for Class 11 Maths Chapter 5 -Complex Numbers and Quadratic Equations

NCERT Exemplar Class 11 Maths Solutions for Chapter 5 - Complex Numbers and Quadratic Equations

NCERT Class 11 Maths Notes for Chapter 5 - Complex Numbers and Quadratic Equations

Practice Questions based on the Euler form of a Complex Number

Below are some practice questions designed to help you understand and apply the Euler form of a complex number, including conversions, calculations, and problem-solving techniques.

Euler Form Of Complex Number - Practice Question MCQ

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