Lesson Explainer: Exponential Form of a Complex Number Mathematics

In this explainer, we will learn how to convert a complex number from the algebraic to the exponential form (Euler’s form) and vice versa.

Definition: Exponential Form of a Complex Number

A complex number 𝑧 can be written in the form 𝑧=π‘Ÿπ‘’, where π‘Ÿ is the modulus and πœƒ is the argument expressed in radians.

If we compare the exponential form to the polar form, we find π‘Ÿπ‘’=π‘Ÿ(πœƒ+π‘–πœƒ);cossin canceling π‘Ÿ, we arrive at 𝑒=πœƒ+π‘–πœƒ.cossin

This formula was derived by the famous mathematician Leonhard Euler (pronounced β€œOiler”) and is called Euler’s formula or Euler’s relation. In many ways, it is a remarkable formula linking the exponential function 𝑒, sine, cosine, πœ‹, and the imaginary unit 𝑖. We can visualize Euler’s formula on an Argand diagram as sweeping out a unit circle centered at the origin.

Another way to visualize it is to consider it in three dimensions, where the value of πœƒ is along one axis and the plane perpendicular to the πœƒ-axis is the complex plane.

This three-dimensional visualization is closely related to the electrodynamical concept of circular polarization for waves.

Returning to Euler’s formula, if we set πœƒ=πœ‹, we get 𝑒=βˆ’1.οƒοŽ„

By adding one to both sides, we derive the famous Euler’s identity: 𝑒+1=0.οƒοŽ„

This equation is considered by many to be an example of mathematical beauty because by using three of the most fundamental operations in mathematics (addition, multiplication, and exponentiation) just once, it links the five fundamental constants of mathematics: 0, 1, 𝑒, πœ‹, and 𝑖.

We will begin by considering a method to derive Euler’s relation using power series. We start with the Maclaurin series expansion for 𝑒: 𝑒=1+π‘₯+π‘₯2+π‘₯3+π‘₯4+π‘₯5+π‘₯!6+π‘₯7+β‹―.ο—οŠ¨οŠ©οŠͺ

Substituting π‘₯=π‘–πœƒ into this equation gives 𝑒=1+π‘–πœƒ+π‘–πœƒ2+π‘–πœƒ3+π‘–πœƒ4+π‘–πœƒ5+π‘–πœƒ6+π‘–πœƒ7+β‹―.οƒοΌοŠ¨οŠ¨οŠ©οŠ©οŠͺοŠͺ

Recall that the integer powers of 𝑖 form a cycle for an integer 𝑛 as follows: 𝑖=1,𝑖=𝑖,𝑖=βˆ’1,𝑖=βˆ’π‘–.οŠͺοŠͺοŠͺοŠͺ

Hence, we have 𝑒=1+π‘–πœƒβˆ’πœƒ2βˆ’π‘–πœƒ3+πœƒ4+π‘–πœƒ5βˆ’πœƒ6βˆ’π‘–πœƒ7+β‹―.οƒοΌοŠ¨οŠ©οŠͺ

Gathering the real and imaginary parts separately, we have 𝑒=ο€Ώ1βˆ’πœƒ2+πœƒ4βˆ’πœƒ6+⋯+π‘–ο€Ώπœƒβˆ’πœƒ3+πœƒ5βˆ’πœƒ7+⋯.οƒοΌοŠ¨οŠͺ

The Maclaurin series of sine and cosine are sincosπœƒ=πœƒβˆ’πœƒ3+πœƒ5βˆ’πœƒ7+β‹―,πœƒ=1βˆ’πœƒ2+πœƒ4βˆ’πœƒ6+β‹―.οŠͺ

Hence, we can see 𝑒=πœƒ+π‘–πœƒ.cossin

To convert a complex number from algebraic to exponential form, we use a very similar technique as that used for converting between algebraic and trigonometric forms. A summary of how to do this is in the box below.

How To: Converting a Complex Number from Algebraic Form to Exponential Form

To convert a complex number in algebraic form, 𝑧=π‘Ž+𝑏𝑖, to exponential form:

  1. Find the modulus, |𝑧|, of the complex number;
  2. Find the argument, arg𝑧, of the complex number;
  3. Write the number in exponential form: 𝑧=π‘Ÿπ‘’, where π‘Ÿ=|𝑧| and πœƒ=𝑧arg.

Example 1: Converting Complex Numbers from Algebraic to Exponential Form

Put the number 𝑧=5√22βˆ’5√62𝑖 in exponential form.


We start by finding the modulus of 𝑧. Substituting the real and imaginary parts into the formula, we have |𝑧|=βˆšπ‘Ž+𝑏=ο„‘ο„£ο„£ο„ ο€Ώ5√22+ο€Ώ5√62.

Simplifying, we get |𝑧|=ο„ž252+752=√50=5√2.

We now find the argument of 𝑧. Notice that since the real part is positive and the imaginary part is negative, 𝑧 lies in the fourth quadrant. Hence, we can find the argument by evaluating the inverse tangent as follows: argarctanarctan(𝑧)=ο€½π‘π‘Žο‰=βŽ›βŽœβŽœβŽβˆ’βŽžβŽŸβŽŸβŽ .√√

Simplifying this fraction gives argarctanarctan(𝑧)=ο€Ώβˆ’βˆš6√2=ο€»βˆ’βˆš3=βˆ’πœ‹3.

Hence, we can express 𝑧 in exponential form as 𝑧=5√2𝑒.οŠ±οƒο‘½οŽ’

Given that the argument is 2πœ‹-periodic, we could equally add 2πœ‹ to the argument and express 𝑧 in exponential form as 𝑧=5√2𝑒.οƒοŽ€ο‘½οŽ’

To convert from exponential form to algebraic form, we rewrite the number in polar form; then, we can convert from this form to algebraic form. In the next example, we will demonstrate this process.

Definition: Converting from Exponential to Algebraic Form

To convert a complex number 𝑧=π‘Ÿπ‘’οƒοΌ to algebraic form, we first convert it to polar form: 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ).cossin

Expanding the brackets and evaluating sine and cosine, we arrive at a complex number in algebraic form:

Using the properties of the modulus and argument, we can derive the rules for multiplication and division of complex numbers in polar form. Recall that for two complex numbers, 𝑧=π‘Ÿπ‘’οŠ§οŠ§οƒοΌοŽ  and 𝑧=π‘Ÿπ‘’οŠ¨οŠ¨οƒοΌοŽ‘, their product can also be expressed in exponential form as 𝑧𝑧=π‘Ÿπ‘’οŠ§οŠ¨οƒοΌ, where π‘Ÿ is the modulus and πœƒ is the argument. Using the properties of the modulus and argument for multiplication, |𝑧𝑧|=|𝑧||𝑧|,(𝑧𝑧)=(𝑧)+(𝑧),argargarg we can see that π‘Ÿ=π‘Ÿπ‘ŸοŠ§οŠ¨ and πœƒ=πœƒ+πœƒοŠ§οŠ¨. Hence, ο€Ήπ‘Ÿπ‘’ο…ο€Ήπ‘Ÿπ‘’ο…=π‘Ÿπ‘Ÿπ‘’.οŠ§οƒοΌοŠ¨οƒοΌοŠ§οŠ¨οƒ()

Likewise, using the properties of the modulus and argument for division, |||𝑧𝑧|||=|𝑧||𝑧|,𝑧𝑧=(𝑧)βˆ’(𝑧),argargarg we can see that π‘Ÿπ‘’π‘Ÿπ‘’=π‘Ÿπ‘Ÿπ‘’.οŠ§οƒοΌοŠ¨οƒοΌοŠ§οŠ¨οƒ()

Notice that if we express the argument as an exponential, we can write π‘Ÿπ‘’=𝑒𝑒=𝑒.()οƒοΌοŽοŠ°οƒοΌlnln

Although the previous derivations might give the impression that all of the rules of exponents apply to complex numbers in general, this is, unfortunately, not true. For example, consider the rule π‘Žπ‘=(π‘Žπ‘), which applies for π‘Ž,𝑏>0. If we let π‘Ž and 𝑏 be negative, this rule can no longer be applied in general; for example, βˆ’1=𝑖=(βˆ’1)(βˆ’1)β‰ ((βˆ’1)(βˆ’1))=1=1.

This just goes to demonstrate that we need to be careful when dealing with complex exponents and bases.

We will now look at some examples using the properties of multiplication and division.

Example 2: Multiplication of Complex Numbers in Exponential Form

Given 𝑧=5π‘’οŠ§οŠ±ο‘½οΌοŽ‘ and 𝑧=6π‘’οŠ¨ο‘½οΌοŽ’, express π‘§π‘§οŠ§οŠ¨ in the form π‘Ž+𝑏𝑖.


Using the multiplicative properties of the exponential form of a complex number, ο€Ήπ‘Ÿπ‘’ο…ο€Ήπ‘Ÿπ‘’ο…=π‘Ÿπ‘Ÿπ‘’,οŠ§οƒοΌοŠ¨οƒοΌοŠ§οŠ¨οƒ() we can write 𝑧𝑧=ο€½5𝑒6𝑒=5Γ—6𝑒=30𝑒.οŠ§οŠ¨οŠ±οŠ±οŠ°οŠ±ο‘½οΌοŽ‘ο‘½οΌοŽ’ο‘½οΌοŽ‘ο‘½οΌοŽ’ο‘½οΌοŽ₯

To convert this to algebraic form, we first express it in polar form as follows: 𝑧𝑧=30ο€»ο€»βˆ’πœ‹6+π‘–ο€»βˆ’πœ‹6.cossin

Expanding the brackets and evaluating sine and cosine, we have 𝑧𝑧=15√3βˆ’15𝑖.

Example 3: Division of Complex Numbers

Given that 𝑍=π‘–βˆš21βˆ’π‘–, write 𝑍 in exponential form.


When faced with a question like this, we have one of two options: we can either convert each number to exponential form and then use their properties to carry out the division, or we can carry out the division with the complex numbers in their current form and then convert the result. Since we have a division to evaluate, it is usually easier to do this with complex numbers in polar form. Hence, we will begin by converting each number to exponential form. Firstly, the numerator is π‘–βˆš2. This is a purely imaginary number; hence, its argument will be πœ‹2. Additionally, its modulus is √2; hence, we can express it in exponential form as √2π‘’ο‘½οΌοŽ‘. As for the denominator, its modulus is 1+(βˆ’1)=√2, and since it lies in the fourth quadrant, its argument can be calculated by evaluating arctanο€Όβˆ’11=βˆ’πœ‹4. Hence, we can express this in exponential form as √2π‘’οŠ±ο‘½οΌοŽ£. Therefore, 𝑍=π‘–βˆš21βˆ’π‘–=√2π‘’βˆš2𝑒.ο‘½οΌοŽ‘ο‘½οΌοŽ£οŠ±

Using the properties of division for complex numbers in exponential form, π‘Ÿπ‘’π‘Ÿπ‘’=π‘Ÿπ‘Ÿπ‘’,οŠ§οƒοΌοŠ¨οƒοΌοŠ§οŠ¨οƒ() we can rewrite this as 𝑍=√2√2𝑒=𝑒.ο‘½οΌοŽ‘ο‘½οΌοŽ£οŽ’ο‘½οΌοŽ£οŠ±ο€ΌοŠ±οˆ

Example 4: Converting Complex Numbers from Exponential to Algebraic Form

Given that 𝑍=π‘’οŠ¨οŠ±οƒοŽ€ο‘½οŽ£, find the algebraic form of 𝑍.


We begin by separating out the real and imaginary parts of the exponent as follows: 𝑍=𝑒=𝑒𝑒.οŠ¨οŠ±οƒοŠ¨οŠ±οƒοŽ€ο‘½οŽ£οŽ€ο‘½οŽ£

We can now convert this to polar form as follows: 𝑍=π‘’ο€Όο€Όβˆ’5πœ‹4+π‘–ο€Όβˆ’5πœ‹4.cossin

Expanding the brackets and evaluating sine and cosine give us the algebraic form of 𝑍 as follows: 𝑍=βˆ’βˆš22𝑒+√22𝑒𝑖.

We can also use Euler’s formula to express sine and cosine in terms of the exponential function as the next example will demonstrate.

Example 5: Relationship between Sine, Cosine, and the Exponential Function

Use Euler’s formula 𝑒=πœƒ+π‘–πœƒοƒοΌcossin to express sinπœƒ and cosπœƒ in terms of 𝑒 and π‘’οŠ±οƒοΌ.


We begin by using Euler’s formula to express π‘’οŠ±οƒοΌ in terms of sine and cosine: 𝑒=𝑒=(βˆ’πœƒ)+𝑖(βˆ’πœƒ).οŠ±οƒοΌοƒ()cossin

Using the even/odd identities for sine and cosine, sinsincoscos(βˆ’πœƒ)=βˆ’(πœƒ),(βˆ’πœƒ)=(πœƒ), we can rewrite this as


Adding this to Euler’s formula, we have 𝑒+𝑒=πœƒ+π‘–πœƒ+πœƒβˆ’π‘–πœƒ=2πœƒ.οƒοΌοŠ±οƒοΌcossincossincos

Dividing by 2, we get cosπœƒ=12𝑒+𝑒.οƒοΌοŠ±οƒοΌ

Similarly, we can derive a formula by sine by considering the difference of equation (1) with Euler’s formula as follows: π‘’βˆ’π‘’=πœƒ+π‘–πœƒβˆ’(πœƒβˆ’π‘–πœƒ)=2π‘–πœƒ.οƒοΌοŠ±οƒοΌcossincossinsin

Dividing by 2𝑖, we arrive at sinπœƒ=12π‘–ο€Ήπ‘’βˆ’π‘’ο….οƒοΌοŠ±οƒοΌ

We will now consider a number of examples where we can use the properties of complex numbers in exponential form to solve problems.

Example 6: Solving Equations Involving Complex Numbers in Exponential Form

Given that π‘Žπ‘’+𝑏𝑒=(2πœƒ)βˆ’5𝑖(2πœƒ)οŠ¨οƒοΌοŠ±οŠ¨οƒοΌcossin, where π‘Žβˆˆβ„ and π‘βˆˆβ„, find π‘Ž and 𝑏.


To solve a problem like this, we want to get the left- and right-hand sides into the same forms. Currently, we have complex numbers in exponential form on the left and a complex number in algebraic form including sine and cosine on the right. We should, therefore, convert the complex number in exponential form to algebraic form. We start by expressing them in polar form: π‘Žπ‘’+𝑏𝑒=π‘Ž(2πœƒ+𝑖2πœƒ)+𝑏((βˆ’2πœƒ)+𝑖(βˆ’2πœƒ)).οŠ¨οƒοΌοŠ±οŠ¨οƒοΌcossincossin

Using the even/odd trigonometric identities, sinsincoscos(βˆ’πœƒ)=βˆ’(πœƒ),(βˆ’πœƒ)=(πœƒ), we can rewrite this as π‘Žπ‘’+𝑏𝑒=π‘Ž(2πœƒ+𝑖2πœƒ)+𝑏(2πœƒβˆ’π‘–2πœƒ).οŠ¨οƒοΌοŠ±οŠ¨οƒοΌcossincossin

Expanding the brackets and gathering like terms, we can express this as π‘Žπ‘’+𝑏𝑒=(π‘Ž+𝑏)2πœƒ+(π‘Žβˆ’π‘)𝑖2πœƒ.οŠ¨οƒοΌοŠ±οŠ¨οƒοΌcossin

We can now equate this with the right-hand side to get (π‘Ž+𝑏)2πœƒ+(π‘Žβˆ’π‘)𝑖2πœƒ=(2πœƒ)βˆ’5𝑖(2πœƒ).cossincossin

Since we know that π‘Ž,π‘βˆˆβ„, we can equate the real and imaginary parts to get the simultaneous equations π‘Ž+𝑏=1,π‘Žβˆ’π‘=βˆ’5.

Adding these equations together, we get 2π‘Ž=βˆ’4. Hence, π‘Ž=βˆ’2. Substituting this back into one of the equations yields 𝑏=3.

Example 7: Properties of Complex Numbers in Exponential Form

Given 𝑧=βˆ’3√3βˆ’3π‘–οŠ§, Imο€Ύπ‘§π‘§οŠ=0, and |||𝑧𝑧|||=3|𝑧|, find all possible values of π‘§οŠ¨, expressing them in exponential form.


We begin by considering what each equation tells us about the value of π‘§οŠ¨. Firstly, consider the equation |||𝑧𝑧|||=3|𝑧|.

Using the properties of the modulus, we can rewrite the left-hand side as |||𝑧𝑧|||=|𝑧|||𝑧||=|𝑧||𝑧|.

Setting this equal to the right-hand side, we have |𝑧||𝑧|=3|𝑧|.

By dividing both sides by |𝑧| and multiplying by |𝑧|, we get 1=3|𝑧|.

Hence, |𝑧|=1√3.

We now consider the equation Imο€Ύπ‘§π‘§οŠ=0.

What does this tell us? It tells us that π‘§π‘§οŠ§οŠ¨οŠ¨ is a real number. Consequently, its argument, which we will denote πœ‘, is either 0 (for a positive real number) or πœ‹ (for a negative real number). Hence, we can write πœ‘=ο€Ύπ‘§π‘§οŠ.arg

Using the properties of the argument, we can rewrite this as πœ‘=(𝑧)βˆ’ο€Ήπ‘§ο…=(𝑧)βˆ’2(𝑧).argargargarg

Rearranging, we get an expression for the argument of π‘§οŠ¨: argarg(𝑧)=12(πœ‘βˆ’(𝑧)).

We now find the argument of π‘§οŠ§. Since π‘§οŠ§ has negative real and imaginary parts, it lies in the third quadrant. Hence, its argument argarctanarctan(𝑧)=ο€Ώβˆ’3βˆ’3√3ο‹βˆ’πœ‹=ο€Ώ1√3ο‹βˆ’πœ‹=πœ‹6βˆ’πœ‹=βˆ’5πœ‹6.

Therefore, we can write arg(𝑧)=12ο€Όπœ‘βˆ’5πœ‹6.

Now we consider the two cases of πœ‘=0 and πœ‘=πœ‹. When πœ‘=0, we have arg(𝑧)=12ο€Όβˆ’5πœ‹6=βˆ’5πœ‹12.

Hence, we can express π‘§οŠ¨ in exponential form as 𝑧=1√3𝑒.οŠ¨οŠ±οƒοŽ€ο‘½οŽ οŽ‘

When πœ‘=πœ‹, we have arg(𝑧)=12ο€Όπœ‹βˆ’5πœ‹6=πœ‹12.

Hence, we can express π‘§οŠ¨ in exponential form as 𝑧=1√3𝑒.οŠ¨οƒο‘½οŽ οŽ‘

Hence, the two possible values of π‘§οŠ¨ are 𝑧=1√3𝑒1√3𝑒.οŠ¨οŠ±οƒοƒοŽ€ο‘½οŽ οŽ‘ο‘½οŽ οŽ‘or

Example 8: Working with Complex Numbers in Exponential Form

Find the numerical value of 𝑒+π‘’οŽ οŽ ο‘½οŽ₯οŽ οŽ ο‘½οŽ₯οƒοŠ±οƒ.


We could approach this problem by converting each number to algebraic form and working through everything. However, we can save ourselves some calculation by being able to recognize a complex conjugate pair when presented in exponential form. Recall that, for any complex number 𝑧, argarg(𝑧)=βˆ’ο€Ήπ‘§ο….

Hence, (π‘Ÿπ‘’)=π‘Ÿπ‘’οƒοΌοŠ±οƒοΌ. Therefore, we have the sum of a complex number and its conjugate. Now if we recall the properties of complex conjugates (i.e., for any complex number, 𝑧+𝑧=2(𝑧)Re), we can simplify our expression: 𝑒+𝑒=2𝑒.οŽ οŽ ο‘½οŽ₯οŽ οŽ ο‘½οŽ₯οŽ οŽ ο‘½οŽ₯οƒοŠ±οƒοƒRe

From Euler’s formula, we know that the real part of 𝑒=πœƒοƒοΌcos. Hence, 𝑒+𝑒=2ο€Ό11πœ‹6=2√32=√3.οŽ οŽ ο‘½οŽ₯οŽ οŽ ο‘½οŽ₯οƒοŠ±οƒcos

Key Points

  • We can express a complex number 𝑧 in exponential form as 𝑧=π‘Ÿπ‘’, where π‘Ÿ is the modulus and πœƒ is the argument expressed in radians.
  • Working with numbers in exponential form can simplify calculations involving multiplication, division, and powers.
  • Using Euler’s formula, we can derive expressions for trigonometric functions such as cossinπœƒ=12𝑒+𝑒,πœƒ=12π‘–ο€Ήπ‘’βˆ’π‘’ο….οƒοΌοŠ±οƒοΌοƒοΌοŠ±οƒοΌ
  • For two complex numbers 𝑧=π‘Ÿπ‘’οŠ§οŠ§οƒοΌοŽ  and 𝑧=π‘Ÿπ‘’οŠ¨οŠ¨οƒοΌοŽ‘, 𝑧𝑧=π‘Ÿπ‘Ÿπ‘’,𝑧𝑧=π‘Ÿπ‘Ÿπ‘’.οŠ§οŠ¨οŠ§οŠ¨οƒ()οŠ§οŠ¨οŠ§οŠ¨οƒ()

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