What is Euler's formula theorem?

Euler's formula states: e^(iӨ) = cosӨ + i sinӨ It expresses complex numbers in exponential form and is known as Euler’s theorem in polar complex representation.

How do you prove Euler's theorem?

Expand e^(iӨ) using the Taylor series and separate the real and imaginary parts. You’ll get \cos x and i \sin x, respectively.

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Euler’s Formula and De Moivre’s Theorem

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Euler’s Formula and De Moivre’s Theorem

Q: What is the difference between Euler’s formula and De Moivre’s theorem?

Euler’s formula relates exponentials to trigonometric functions, while De Moivre’s theorem helps calculate powers and roots of complex numbers using Euler’s form.

Euler’s Formula and De Moivre’s Theorem are powerful tools in complex number theory. Euler’s Formula, $e^{i θ} = cos θ + i s in θ$ , bridges exponential and trigonometric forms, while De Moivre’s Theorem, $(cos θ + i s in θ)^{n} = cos (n θ) + i s in (n θ)$ , 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: $e^{i x} = cos x + i s in x$

Where:

e is the base of the natural logarithm,
i is the imaginary unit $(- 1)$
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: $e^{i θ} = cos θ + i s in θ$

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:

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

Breaking this into real and imaginary parts:

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

Hence,

$e^{i x} = cos x + i s in x$

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 $z = cos θ + i s in θ$ , then:

$(cos θ + i s in θ)^{n} = cos (n θ) + i s in (n θ)$

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

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:

$(cos θ + i s in θ)^{n} = (e^{i θ})^{n} = e^{in θ} = cos (n θ) + i s in (n θ)$

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?

Feature	Euler’s Formula	De Moivre’s Theorem
Expression	$e^{i θ} = cos θ + i s in θ$	$(cos θ + i s in θ)^{n} = cos (n θ) + i s in (n θ)$
Focus	Describes exponential form of complex numbers	Calculates powers/roots of complex numbers
Based On	Taylor series	Built upon Euler’s formula
Application	Conversion & analysis	Trigonometry, solving complex equations

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

Start with:

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

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 $cos θ + i s in θ$ in exponential form.

Solution:

Using Euler's formula:

$e^{i θ} = cos θ + i s in θ$

$\Rightarrow C os θ + i S in θ = e^{i θ}$

Example 2: Find $(cos \frac{π}{3} + i s in \frac{π}{3})^{5}$

Solution:

Using De Moivre’s Theorem:

$(cos θ + i s in θ)^{n} = cos (n θ) + i s in (n θ)$

$= cos (5. \frac{π}{3}) + i s in (5. \frac{π}{3})$

$= cos (\frac{5 π}{3}) + i s in (\frac{5 π}{3})$

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

Example 3: Prove that $(cos θ + i s in θ)^{n} + (cos θ - i s in θ)^{n}$ is real.

Solution:

Using De Moivre’s Theorem:

$(cos θ + i s in θ)^{n} = cos (n θ) + i s in (n θ)$

$(cos θ - i s in θ)^{n} - cos (n θ) - i s in (n θ)$

Adding:

$= 2 cos (n θ) \Rightarrow A Real Number$

Example 4: Find all cube roots of $8 (cos \frac{π}{3} + i s in \frac{π}{3})$

Solution:

Use De Moivre’s Theorem for nth roots:

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

Given, $r = 8, θ = \frac{π}{3}, n = 3$

$38 = 2 \Rightarrow Roots:$

$2 [cos (\frac{π}{9}) + i sin (\frac{π}{9})]$
$2 [cos (\frac{7 π}{9}) + i sin (\frac{7 π}{9})]$
$2 [cos (\frac{13 π}{9}) + i sin (\frac{13 π}{9})]$

Example 5: Prove $e^{iπ} + 1 = 0$

Solution:

Using Euler’s Formula:

$e^{iπ} = cos π + i s inπ$

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

$\Rightarrow e^{iπ} + 1 = 0$

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

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

Solution:

Euler’s formulas:

$e^{i θ} = cos θ + i s in θ$ and $e^{- i θ} = cos θ - i s in θ$

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

Thus, $C os θ = \frac{e ^{i θ} + e ^{- i θ}}{2}$

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 $z = c i s θ = cos θ + i s in θ$

$z^{3} = c i s (30) = 1 = c i s (0 + 2 kπ), f or k = 0, 1, 2$

Then,

$3 θ = 2 kπ \Rightarrow θ = \frac{2 kπ}{3}$

Thus, roots are:

$z_{0} = c i s 0 = 1, z_{1} = c i s (\frac{2 π}{3}), z_{2} = c i s (\frac{4 π}{3})$

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

Solution:
By De Moivre's theorem:

$(cos θ + i s in θ)^{3} = cos (3 θ) + i s in (3 θ)$

Expand LHS using binomial theorem:

$= co s^{3} θ + 3 i co s^{2} θ s in θ - 3 cos θ s i n^{2} θ - i s i n^{3} θ$

Group real and imaginary parts:

Real part = $co s^{3} θ - 3 cos θ s i n^{2} θ$

But $s i n^{2} θ = 1 - co s^{2} θ$ , so:

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

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

Example 9: Evaluate 5 $(cos \frac{π}{5} + i s in \frac{π}{5})^{5}$

Solution: By De Moivre’s Theorem:

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

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

Solution:

From Euler’s Formula:

$e^{i θ} = cos θ + i s in θ, e^{- i θ} = cos θ - i s in θ$

Add:

$e^{i θ} + e^{- i θ} = 2 cos θ$

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

Subtract:

$e^{i θ} - e^{- i θ} = 2 i sin θ$

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

Example 11: Find $(cos \frac{π}{4} + i s in \frac{π}{4})^{3}$

Solution:

Using De Moivre’s Theorem:

$= cos (3. \frac{π}{3}) + i s in (3. \frac{π}{4})$

$= cos \frac{3 π}{4} + i s in \frac{3 π}{4}$

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

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

Solution:

Cube roots of unity:

$z = 1 = e^{2 kiπ /3} for k=0,1,2$

$\Rightarrow z_{1} = 1, z_{2} = ω, z_{3} = ω^{2}$

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

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

Evaluate $(cos \frac{π}{6} + i s in \frac{π}{6})^{4}$ using De Moivre’s Theorem.
Express $cos \frac{π}{2} + i s in \frac{π}{2}$ in exponential form.
Find the 4th roots of $16 (cos \frac{π}{4} + i s in \frac{π}{4})$ .
If $z = e^{i θ}$ , show that $z^{n} + \overset{z}{ˉ}^{n}$ is real.
Prove: $(cos x + i s in x)^{n} - (cos x - i s in x)^{n} = 2 i s in (n x)$

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

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

Ans: $(C os θ + i s in θ)^{n} = cos (n θ) + i s in (n θ)$

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

Ans: Using Euler’s form:

$z = e^{i θ}$

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

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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, $e^{i θ} = cos θ + i s in θ$ , bridges exponential and trigonometric forms, while De Moivre’s Theorem, $(cos θ + i s in θ)^{n} = cos (n θ) + i s in (n θ)$ , 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: $e^{i x} = cos x + i s in x$

Where:

e is the base of the natural logarithm,
i is the imaginary unit $(- 1)$
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: $e^{i θ} = cos θ + i s in θ$

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:

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

Breaking this into real and imaginary parts:

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

Hence,

$e^{i x} = cos x + i s in x$

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 $z = cos θ + i s in θ$ , then:

$(cos θ + i s in θ)^{n} = cos (n θ) + i s in (n θ)$

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

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:

$(cos θ + i s in θ)^{n} = (e^{i θ})^{n} = e^{in θ} = cos (n θ) + i s in (n θ)$

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?

Feature	Euler’s Formula	De Moivre’s Theorem
Expression	$e^{i θ} = cos θ + i s in θ$	$(cos θ + i s in θ)^{n} = cos (n θ) + i s in (n θ)$
Focus	Describes exponential form of complex numbers	Calculates powers/roots of complex numbers
Based On	Taylor series	Built upon Euler’s formula
Application	Conversion & analysis	Trigonometry, solving complex equations

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

Start with:

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

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 $cos θ + i s in θ$ in exponential form.

Solution:

Using Euler's formula:

$e^{i θ} = cos θ + i s in θ$

$\Rightarrow C os θ + i S in θ = e^{i θ}$

Example 2: Find $(cos \frac{π}{3} + i s in \frac{π}{3})^{5}$

Solution:

Using De Moivre’s Theorem:

$(cos θ + i s in θ)^{n} = cos (n θ) + i s in (n θ)$

$= cos (5. \frac{π}{3}) + i s in (5. \frac{π}{3})$

$= cos (\frac{5 π}{3}) + i s in (\frac{5 π}{3})$

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

Example 3: Prove that $(cos θ + i s in θ)^{n} + (cos θ - i s in θ)^{n}$ is real.

Solution:

Using De Moivre’s Theorem:

$(cos θ + i s in θ)^{n} = cos (n θ) + i s in (n θ)$

$(cos θ - i s in θ)^{n} - cos (n θ) - i s in (n θ)$

Adding:

$= 2 cos (n θ) \Rightarrow A Real Number$

Example 4: Find all cube roots of $8 (cos \frac{π}{3} + i s in \frac{π}{3})$

Solution:

Use De Moivre’s Theorem for nth roots:

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

Given, $r = 8, θ = \frac{π}{3}, n = 3$

$38 = 2 \Rightarrow Roots:$

$2 [cos (\frac{π}{9}) + i sin (\frac{π}{9})]$
$2 [cos (\frac{7 π}{9}) + i sin (\frac{7 π}{9})]$
$2 [cos (\frac{13 π}{9}) + i sin (\frac{13 π}{9})]$

Example 5: Prove $e^{iπ} + 1 = 0$

Solution:

Using Euler’s Formula:

$e^{iπ} = cos π + i s inπ$

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

$\Rightarrow e^{iπ} + 1 = 0$

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

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

Solution:

Euler’s formulas:

$e^{i θ} = cos θ + i s in θ$ and $e^{- i θ} = cos θ - i s in θ$

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

Thus, $C os θ = \frac{e ^{i θ} + e ^{- i θ}}{2}$

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 $z = c i s θ = cos θ + i s in θ$

$z^{3} = c i s (30) = 1 = c i s (0 + 2 kπ), f or k = 0, 1, 2$

Then,

$3 θ = 2 kπ \Rightarrow θ = \frac{2 kπ}{3}$

Thus, roots are:

$z_{0} = c i s 0 = 1, z_{1} = c i s (\frac{2 π}{3}), z_{2} = c i s (\frac{4 π}{3})$

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

Solution:
By De Moivre's theorem:

$(cos θ + i s in θ)^{3} = cos (3 θ) + i s in (3 θ)$

Expand LHS using binomial theorem:

$= co s^{3} θ + 3 i co s^{2} θ s in θ - 3 cos θ s i n^{2} θ - i s i n^{3} θ$

Group real and imaginary parts:

Real part = $co s^{3} θ - 3 cos θ s i n^{2} θ$

But $s i n^{2} θ = 1 - co s^{2} θ$ , so:

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

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

Example 9: Evaluate 5 $(cos \frac{π}{5} + i s in \frac{π}{5})^{5}$

Solution: By De Moivre’s Theorem:

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

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

Solution:

From Euler’s Formula:

$e^{i θ} = cos θ + i s in θ, e^{- i θ} = cos θ - i s in θ$

Add:

$e^{i θ} + e^{- i θ} = 2 cos θ$

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

Subtract:

$e^{i θ} - e^{- i θ} = 2 i sin θ$

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

Example 11: Find $(cos \frac{π}{4} + i s in \frac{π}{4})^{3}$

Solution:

Using De Moivre’s Theorem:

$= cos (3. \frac{π}{3}) + i s in (3. \frac{π}{4})$

$= cos \frac{3 π}{4} + i s in \frac{3 π}{4}$

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

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

Solution:

Cube roots of unity:

$z = 1 = e^{2 kiπ /3} for k=0,1,2$

$\Rightarrow z_{1} = 1, z_{2} = ω, z_{3} = ω^{2}$

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

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

Evaluate $(cos \frac{π}{6} + i s in \frac{π}{6})^{4}$ using De Moivre’s Theorem.
Express $cos \frac{π}{2} + i s in \frac{π}{2}$ in exponential form.
Find the 4th roots of $16 (cos \frac{π}{4} + i s in \frac{π}{4})$ .
If $z = e^{i θ}$ , show that $z^{n} + \overset{z}{ˉ}^{n}$ is real.
Prove: $(cos x + i s in x)^{n} - (cos x - i s in x)^{n} = 2 i s in (n x)$

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

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

Ans: $(C os θ + i s in θ)^{n} = cos (n θ) + i s in (n θ)$

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

Ans: Using Euler’s form:

$z = e^{i θ}$

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

Frequently Asked Questions

What is Euler's formula theorem?

How do you prove Euler's theorem?

What is the difference between Euler’s formula and De Moivre’s theorem?

Frequently Asked Questions

What is Euler's formula theorem?

How do you prove Euler's theorem?

What is the difference between Euler’s formula and De Moivre’s theorem?

Join ALLEN!

Euler’s Formula and De Moivre’s Theorem

1.0What is Euler’s Formula?

2.0What is Euler’s Theorem?

3.0How to Prove Euler’s Theorem

4.0De Moivre’s Theorem

De Moivre’s Formula

5.0Applications of De Moivre's Theorem:

6.0Euler’s Formula and De Moivre’s Theorem

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?

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

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

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

Table of Contents

Join ALLEN!

Euler’s Formula and De Moivre’s Theorem

1.0What is Euler’s Formula?

2.0What is Euler’s Theorem?

3.0How to Prove Euler’s Theorem

4.0De Moivre’s Theorem

De Moivre’s Formula

5.0Applications of De Moivre's Theorem:

6.0Euler’s Formula and De Moivre’s Theorem

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?

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

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

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

Table of Contents