Euler’s Formula and De Moivre’s Theorem

Euler’s Formula and De Moivre’s Theorem are powerful tools in complex number theory. Euler’s Formula, $e^{i\theta }=cos\theta +isin\theta$, bridges exponential and trigonometric forms, while De Moivre’s Theorem, $(cos\theta +isin\theta {)}^n=cos(n\theta )+isin(n\theta )$, simplifies powers and roots of complex numbers. These theorems play a crucial role in advanced mathematics, particularly in JEE-level problems involving polar forms, complex roots, and trigonometric identities.

1.0What is Euler’s Formula?

Euler’s formula is a fundamental identity in complex analysis that connects the exponential function with trigonometric functions. It states: $e^{ix}=cosx+isinx$

Where:

• e is the base of the natural logarithm,
• i is the imaginary unit $(\sqrt{-1})$
• x is a real number (often an angle in radians).

2.0What is Euler’s Theorem?

Sometimes, Euler’s formula is referred to as Euler’s theorem, especially when applied to complex numbers in exponential form. In this context:

Euler’s Theorem: $e^{i\theta }=cos\theta +isin\theta$

This is often used to express complex numbers in polar or exponential form.

3.0How to Prove Euler’s Theorem

To prove Euler's theorem, we use the Taylor series expansion of $e^x$, cos x, and sin x:

Step-by-step:

 $e^{ix}=1+ix+\frac{(ix{)}^2}{2!}+\frac{(ix{)}^3}{3!}+\frac{(ix{)}^4}{4!}+...$

Breaking this into real and imaginary parts:

• Real part: $1-\frac{x^2}{2!}+\frac{x^4}{4!}+..$
• Imaginary part: $ix-\frac{i{x}^3}{3!}+\frac{i{x}^5}{5!}-....=isinx$

Hence,

$e^{ix}=cosx+isinx$

This completes the proof of Euler’s theorem.

4.0De Moivre’s Theorem

De Moivre’s Theorem is a powerful tool to compute powers and roots of complex numbers in polar form.

De Moivre’s Formula

If $z=cos\theta +isin\theta$, then:

$(cos\theta +isin\theta {)}^n=cos(n\theta )+isin(n\theta )$

Or using Euler’s form: $(e^{i\theta }{)}^n=e^{in\theta }$

5.0Applications of De Moivre's Theorem:

• Finding powers of complex numbers
• Finding nth roots of complex numbers
• Trigonometric identities
• Solving complex number equations

6.0Euler’s Formula and De Moivre’s Theorem

The two formulas are tightly connected.

Using Euler's formula:

$(cos\theta +isin\theta {)}^n=(e^{i\theta }{)}^n=e^{in\theta }=cos(n\theta )+isin(n\theta )$

Thus, De Moivre’s Theorem is a direct consequence of Euler’s formula.

7.0What is the Difference Between Euler’s Formula and De Moivre’s Theorem?

8.0How Do You Use Euler's Formula to Prove De Moivre’s Formula?

Start with:

$z=e^{i\theta }\Rightarrow z^n=(e^{i\theta }{)}^n=e^{in\theta }=cos(n\theta )+isin(n\theta )$

Which matches the RHS of De Moivre’s Theorem.

Thus, Euler’s formula is used to derive De Moivre’s formula.

9.0Solved Examples on Euler’s Formula and De Moivre’s Theorem

Example 1: Express $cos\theta +isin\theta$ in exponential form.

Solution:

Using Euler's formula:

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

$\Rightarrow Cos\theta +iSin\theta =e^{i\theta }$

Example 2: Find $(cos\frac{\pi }{3}+isin\frac{\pi }{3}{)}^5$

Solution:

Using De Moivre’s Theorem:

$(cos\theta +isin\theta {)}^n=cos(n\theta )+isin(n\theta )$

$=cos(5.\frac{\pi }{3})+isin(5.\frac{\pi }{3})$

$=cos(\frac{5\pi }{3})+isin(\frac{5\pi }{3})$

$=\frac{1}{2}-i.\frac{\sqrt{3}}{2}$

Example 3: Prove that $(cos\theta +isin\theta {)}^n+(cos\theta -isin\theta {)}^n$ is real.

Solution:

Using De Moivre’s Theorem:

$(cos\theta +isin\theta {)}^n=cos(n\theta )+isin(n\theta )$

$(cos\theta -isin\theta {)}^n-cos(n\theta )-isin(n\theta )$

Adding:

$=2cos(n\theta )\Rightarrow \text{A Real Number}$

Example 4: Find all cube roots of $8(cos\frac{\pi }{3}+isin\frac{\pi }{3})$

Solution:

Use De Moivre’s Theorem for nth roots:

$z=r^{1/n}\left[\cos \left(\frac{\theta +2k\pi }{n}\right)+i\sin \left(\frac{\theta +2k\pi }{n}\right)\right],k=0,1,2$

Given, $r=8,\theta =\frac{\pi }{3},n=3$

$\sqrt[3]{8}=2\Rightarrow \text{Roots:}$

1. $2\left[\cos \left(\frac{\pi }{9}\right)+i\sin \left(\frac{\pi }{9}\right)\right]$
2. $2\left[\cos \left(\frac{7\pi }{9}\right)+i\sin \left(\frac{7\pi }{9}\right)\right]$
3. $2\left[\cos \left(\frac{13\pi }{9}\right)+i\sin \left(\frac{13\pi }{9}\right)\right]$

Example 5: Prove $e^{i\pi }+1=0$

Solution:

Using Euler’s Formula:

$e^{i\pi }=cos\pi +isin\pi$

$\Rightarrow e^{i\pi }=-1+i.0=-1$

$\Rightarrow e^{i\pi }+1=0$

Known as Euler’s Identity — one of the most beautiful equations in mathematics!

Example 6: Express $cos\theta$ in terms of $e^{i\theta }$ and $e^{-i\theta }$ using Euler’s Formula.

Solution:

Euler’s formulas:

$e^{i\theta }=cos\theta +isin\theta$ and $e^{-i\theta }=cos\theta -isin\theta$

Add both equations: $e^{i\theta }+e^{-i\theta }=2cos\theta$

Thus, $Cos\theta =\frac{e^{i\theta }+e^{-i\theta }}{2}$

Example 7: Find all cube roots of unity using De Moivre’s Theorem.

Solution:
We know cube roots of unity are solutions of $z^3=1$.

So, let $z=cis\theta =cos\theta +isin\theta$

$z^3=cis(30)=1=cis(0+2k\pi ),fork=0,1,2$ 

Then,

$3\theta =2k\pi \Rightarrow \theta =\frac{2k\pi }{3}$

Thus, roots are:

 $z_0=cis0=1,z_1=cis(\frac{2\pi }{3}),z_2=cis(\frac{4\pi }{3})$

Example 8: Prove that $cos3\theta =4co{s}^3\theta -3cos\theta$using De Moivre’s Theorem.

Solution:
By De Moivre's theorem:

$(cos\theta +isin\theta {)}^3=cos(3\theta )+isin(3\theta )$

Expand LHS using binomial theorem:

$=co{s}^3\theta +3ico{s}^2\theta sin\theta -3cos\theta si{n}^2\theta -isi{n}^3\theta$

Group real and imaginary parts:

Real part = $co{s}^3\theta -3cos\theta si{n}^2\theta$

But $si{n}^2\theta =1-co{s}^2\theta$, so:

$co{s}^3\theta -3cos\theta (1-co{s}^2\theta )=4co{s}^3\theta -3cos\theta$

Hence, $cos(3\theta )=4co{s}^2\theta -3cos\theta$

Example 9: Evaluate 5 $(cos\frac{\pi }{5}+isin\frac{\pi }{5}{)}^5$

Solution: By De Moivre’s Theorem:

${\left(\cos \frac{\pi }{5}+i\sin \frac{\pi }{5}\right)}^5=\cos (\pi )+i\sin (\pi )=-1+i(0)=-1$

Example 10: Prove $\cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}$ and $\sin \theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}$

Solution:

From Euler’s Formula:

$e^{i\theta }=cos\theta +isin\theta ,e^{-i\theta }=cos\theta -isin\theta$

Add:

$e^{i\theta }+e^{-i\theta }=2\cos \theta$

$\Rightarrow \cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}$ 

Subtract:

$e^{i\theta }-e^{-i\theta }=2i\sin \theta$

$\Rightarrow \sin \theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}$ 

Example 11: Find $(cos\frac{\pi }{4}+isin\frac{\pi }{4}{)}^3$

Solution: 

Using De Moivre’s Theorem:

$=cos(3.\frac{\pi }{3})+isin(3.\frac{\pi }{4})$

$=cos\frac{3\pi }{4}+isin\frac{3\pi }{4}$

$=-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$

Example 12: Find the cube roots of unity using De Moivre’s Theorem.

Solution: 

Cube roots of unity:

$z=1=e^{2ki\pi /3}\text{for k=0,1,2}$

$\Rightarrow z_1=1,z_2=\omega ,z_3={\omega }^2$ 

Where $\omega =-\frac{1}{2}+\frac{\sqrt{3}}{2}i$

10.0Practice Questions on Euler’s Formula and De Moivre’s Theorem

1. Evaluate $(cos\frac{\pi }{6}+isin\frac{\pi }{6}{)}^4$ using De Moivre’s Theorem.
2. Express $cos\frac{\pi }{2}+isin\frac{\pi }{2}$ in exponential form.
3. Find the 4th roots of $16(cos\frac{\pi }{4}+isin\frac{\pi }{4})$.
4. If $z=e^{i\theta }$, show that $z^n+{\bar{z}}^n$ is real.
5. Prove: $(cosx+isinx{)}^n-(cosx-isinx{)}^n=2isin(nx)$

11.0Sample Questions on Euler’s Formula and De Moivre’s Theorem

Q1. What is the formula for De Moivre's theorem?

Ans: $(Cos\theta +isin\theta {)}^n=cos(n\theta )+isin(n\theta )$

Q2. How do you use Euler's formula to prove De Moivre’s formula?

Ans: Using Euler’s form:

 $z=e^{i\theta }$

$\Rightarrow z^n=e^{in\theta }=cos(n\theta )+isin(n\theta )$