Lesson Explainer: Exponential Form of a Complex Number | Nagwa Lesson Explainer: Exponential Form of a Complex Number | Nagwa

Lesson Explainer: Exponential Form of a Complex Number Mathematics • Third Year of Secondary School

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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.

Let us begin by recalling the polar form of a complex number. If a complex number 𝑧 has modulus π‘Ÿ and argument πœƒ, then the polar form of the complex number is 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ).cossin

This gives us a way to represent the complex number using only the modulus and the argument. The exponential form of a complex number is an alternative and simpler way to represent a complex number using its modulus and argument. To obtain the exponential form of a complex number, we need to understand how to simplify the expression cossinπœƒ+π‘–πœƒ, which appears in the polar form. For this purpose, we introduce a very important formula credited to a Swiss mathematician, Leonhard Euler (1707–1783).

Formula: Euler’s Formula

For any real number πœƒ, 𝑒=πœƒ+π‘–πœƒ.cossin

This equation is known as Euler’s identity or Euler’s formula. This remarkable identity links the exponential function 𝑒, trigonometric functions, 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.

A particular case of Euler’s formula is when πœƒ=πœ‹, we have 𝑒=βˆ’1.οƒοŽ„

By adding 1 to both sides, we derive the famous 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 can prove Euler’s formula using Maclaurin series. The Maclaurin or Taylor series of a function allows us to extend a real variable function to take complex-valued inputs, as long as the series converges. We begin 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

Since the left-hand side of Euler’s formula has 𝑒, we substitute π‘₯=π‘–πœƒ into this equation to write 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

Recall that 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

Replacing the power series in the real and imaginary parts of 𝑒 with suitable Maclaurin series, we obtain 𝑒=πœƒ+π‘–πœƒ.cossin

This proves Euler’s formula.

In our first example, we will use Euler’s formula to write sinπœƒ and cosπœƒ in terms of complex exponential functions.

Example 1: 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.

Answer

Recall Euler’s formula: 𝑒=πœƒ+π‘–πœƒ.cossin

We begin by applying Euler’s formula to express π‘’οŠ±οƒοΌ by substituting βˆ’πœƒ in place of πœƒ in the formula above: 𝑒=𝑒=(βˆ’πœƒ)+𝑖(βˆ’πœƒ).οŠ±οƒοΌοƒ()cossin

Recall the even/odd identities for sine and cosine: sinsincoscos(βˆ’πœƒ)=βˆ’(πœƒ),(βˆ’πœƒ)=(πœƒ).

Substituting these identities above, we can rewrite 𝑒=πœƒβˆ’π‘–πœƒ.οŠ±οƒοΌcossin

Now, we have two identities: 𝑒=πœƒ+π‘–πœƒ,𝑒=πœƒβˆ’π‘–πœƒ.οƒοΌοŠ±οƒοΌcossincossin

We can think of these as simultaneous equations where the unknowns are cosπœƒ and sinπœƒ. If we add these two equations, we can remove the sine function and find the expression for cosπœƒ in terms of 𝑒 and π‘’οŠ±οƒοΌ. Also, if we subtract these equations, we can remove the cosine function and find the expression for sinπœƒ in terms of 𝑒 and π‘’οŠ±οƒοΌ.

Let us first find the expression for cosπœƒ in terms of 𝑒 and π‘’οŠ±οƒοΌ by adding the two simultaneous equations: 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

This gives the expression for cosπœƒ in terms of 𝑒 and π‘’οŠ±οƒοΌ.

Next, let us find a similar formula for sinπœƒ by taking the difference of the simultaneous equations: 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

This gives the expression for sinπœƒ in terms of 𝑒 and π‘’οŠ±οƒοΌ.

Hence, we have obtained cossinπœƒ=12𝑒+𝑒,πœƒ=12π‘–ο€Ήπ‘’βˆ’π‘’ο….οƒοΌοŠ±οƒοΌοƒοΌοŠ±οƒοΌ

In the previous example, we applied Euler’s formula to express the sine and cosine functions in terms of complex exponential functions. The primary application of Euler’s formula in this explainer is to convert the polar form of a complex number to the exponential form.

Recall that the polar form of a complex number 𝑧 with modulus π‘Ÿ and argument πœƒ is 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ).cossin

Euler’s formula tells us that the expression inside the parentheses is equal to 𝑒. When we make this substitution, we obtain the exponential form of a complex number.

Definition: Exponential Form of a Complex Number

The exponential form of a nonzero complex number 𝑧 with modulus π‘Ÿ and argument πœƒ is 𝑧=π‘Ÿπ‘’.

The standard range of the argument πœƒ in the exponential form is ]βˆ’πœ‹,πœ‹] radians.

To write the exponential form of a complex number, we need to know its modulus and argument, which are the same characteristics required for the polar form of the complex number. Hence, converting complex numbers between the polar and exponential forms is a simple task.

Let us consider an example where we will convert a complex number from the polar form to the exponential form.

Example 2: Converting Complex Numbers from Polar to Exponential Form

Put 𝑧=4√3ο€Ό5πœ‹6βˆ’π‘–5πœ‹6cossin in exponential form.

Answer

Recall that the exponential form of a nonzero complex number 𝑧 with modulus π‘Ÿ and argument πœƒ is 𝑧=π‘Ÿπ‘’.

We are given the complex number in a form that is close to the polar form. We know that the polar form of a nonzero complex number 𝑧 with modulus π‘Ÿ and argument πœƒ is 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ).cossin

Since we know the polar form of a complex number, we can identify the modulus and argument of the complex number, which then would lead to the exponential form of the complex number.

However, the given form of our complex number has a negative sign in front of the imaginary part, which means that it is not exactly the polar form. We need to first convert this number to the polar form so that we can identify its modulus and argument. For this purpose, we need to recall that the sine function is odd and the cosine function is even. This leads to the identities sinsincoscosπœƒ=βˆ’(βˆ’πœƒ),πœƒ=(βˆ’πœƒ).

Applying these identities to our number, we can write 𝑧=4√3ο€Όο€Όβˆ’5πœ‹6+π‘–ο€Όβˆ’5πœ‹6.cossin

This is the polar form of 𝑧. This means that the modulus of 𝑧 is 4√3, and the argument of 𝑧 is βˆ’5πœ‹6. We recall that the standard range of arguments is ]βˆ’πœ‹,πœ‹], so our argument is within the correct range. This leads to the exponential form 𝑧=4√3𝑒.οŽͺοŽ€ο‘½οŽ₯

In the previous example, we converted a complex number from the polar form to the exponential form. This process is simple since the modulus and argument of a complex number are already provided in the polar form. If we start with the Cartesian form of a complex number, we must first compute the modulus and argument of the complex number in order to obtain the exponential form. Let us consider an example where we convert a complex number from the Cartesian form to the exponential form.

Example 3: Converting Complex Numbers from Cartesian to Exponential Form

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

Answer

Recall that the exponential form of a nonzero complex number 𝑧 with modulus π‘Ÿ and argument πœƒ is 𝑧=π‘Ÿπ‘’.

We start by finding the modulus of 𝑧. Recall that the modulus of a complex number π‘Ž+𝑏𝑖 is given by βˆšπ‘Ž+π‘οŠ¨οŠ¨. Substituting the real part π‘Ž=5√22 and imaginary part 𝑏=βˆ’5√62 into this formula and simplifying, we have |𝑧|=βˆšπ‘Ž+𝑏=ο„‘ο„£ο„£ο„ ο€Ώ5√22+ο€Ώβˆ’5√62=ο„ž252+752=√50=5√2.

This tells us that the modulus of 𝑧 is 5√2.

Let us 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 of an Argand diagram. Recall that the argument of a complex number π‘Ž+𝑏𝑖 in the fourth quadrant is given by arctanπ‘π‘Ž. Hence, argarctanarctanarctanarctan(𝑧)=ο€½π‘π‘Žο‰=βŽ›βŽœβŽœβŽβŽžβŽŸβŽŸβŽ =ο€Ώβˆ’βˆš6√2=ο€»βˆ’βˆš3=βˆ’πœ‹3.√√

This means that the argument of 𝑧 is βˆ’πœ‹3. We recall that the standard range of arguments is ]βˆ’πœ‹,πœ‹], so our argument is within the correct range.

Using the modulus π‘Ÿ=5√2 and the argument πœƒ=βˆ’πœ‹3, we can express 𝑧 in the exponential form as 𝑧=5√2𝑒.οŽͺο‘½οŽ’οƒ

In the previous example, we converted a complex number in the Cartesian form to the exponential form. We now consider the converse process. To convert a complex number in the exponential form to the Cartesian form, we need to first convert it to the polar form. We can then convert the polar form of the complex number to the Cartesian form. In the next example, we will demonstrate this process.

Example 4: Converting Complex Numbers from Exponential to Cartesian Form

Put 𝑧=5√3π‘’ο‘½οŽ’οƒ in Cartesian form.

Answer

We are given the exponential form of a complex number. To convert a complex number in the exponential form to the Cartesian form, we need to first convert it to the polar form. We can then convert the polar form of the complex number to the Cartesian form.

Recall that the exponential form of a nonzero complex number 𝑧 with modulus π‘Ÿ and argument πœƒ is 𝑧=π‘Ÿπ‘’.

We also recall that the polar form of a nonzero complex number 𝑧 with modulus π‘Ÿ and argument πœƒ is 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ).cossin

We can see that the exponential and polar forms share the same parameters π‘Ÿ and πœƒ, which makes the conversion simpler.

From the given exponential form, we obtain modulus 5√3 and argument πœ‹3. Hence, the polar form of 𝑧 is 𝑧=5√3ο€»πœ‹3+π‘–πœ‹3.cossin

To convert the polar form of a complex number to the Cartesian form, we need to multiply through the parentheses and evaluate the trigonometric ratios. Since πœ‹3 is a special angle, we recall the trigonometric ratios cossinπœ‹3=12,πœ‹3=√32.

Substituting these values into the polar form of 𝑧 and simplifying, we obtain 𝑧=5√3ο€Ώ12+π‘–βˆš32=5√32+𝑖5√3Γ—βˆš32=5√32+𝑖152.

Hence, the Cartesian form of the given complex number is 𝑧=5√32+𝑖152.

Using the properties of the modulus and argument under multiplication, the multiplication of a pair of nonzero complex numbers is simpler in the polar or exponential form. Let us recall these properties: |𝑧𝑧|=|𝑧||𝑧|,(𝑧𝑧)=(𝑧)+(𝑧).argargarg

This leads to the multiplication rule for complex numbers in the exponential form.

Rule: Multiplication of Complex Numbers in Exponential Form

Let 𝑧=π‘Ÿπ‘’οŠ§οŠ§οƒοΌοŽ  and 𝑧=π‘Ÿπ‘’οŠ¨οŠ¨οƒοΌοŽ‘ be complex numbers in the exponential form. The product π‘§π‘§οŠ§οŠ¨ in the exponential form is ο€Ήπ‘Ÿπ‘’ο…ο€Ήπ‘Ÿπ‘’ο…=π‘Ÿπ‘Ÿπ‘’.οŠ§οƒοΌοŠ¨οƒοΌοŠ§οŠ¨οƒ()

If we recall the law of exponents that states π‘Žπ‘Ž=π‘Ž,π‘Ž>0,ο—ο˜ο—οŠ°ο˜forany then the multiplication rule for the exponential form is very intuitive. In fact, this rule states that the law of exponents still holds when the exponents are complex numbers. This is true as long as the base of the exponent is a positive real number.

Let us consider an example where we will multiply a pair of complex numbers in the exponential form.

Example 5: Multiplication of Complex Numbers in Exponential Form

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

Answer

Recall the multiplication rule for complex numbers in the exponential form, ο€Ήπ‘Ÿπ‘’ο…ο€Ήπ‘Ÿπ‘’ο…=π‘Ÿπ‘Ÿπ‘’.οŠ§οƒοΌοŠ¨οƒοΌοŠ§οŠ¨οƒ()

Since we are given both complex numbers π‘§οŠ§ and π‘§οŠ¨ in the exponential form, we can apply this rule with the values π‘Ÿ=5,π‘Ÿ=6,πœƒ=βˆ’πœ‹2,πœƒ=πœ‹3.

Substituting these values into the multiplication rule, we obtain 𝑧𝑧=ο€½5𝑒6𝑒=5Γ—6𝑒=30𝑒.οŽͺο‘½οΌοŽ‘ο‘½οΌοŽ’οŽͺο‘½οΌοŽ‘ο‘½οΌοŽ’οŽͺοŽ₯

This leads to the polar form of the complex number π‘§π‘§οŠ§οŠ¨.

To convert a complex number in the exponential form to the Cartesian form, we need to first convert it to the polar form. We can then convert the polar form of the complex number to the Cartesian form.

Recall that the exponential form of a nonzero complex number 𝑧 with modulus π‘Ÿ and argument πœƒ is 𝑧=π‘Ÿπ‘’.

We also recall that the polar form of a nonzero complex number 𝑧 with modulus π‘Ÿ and argument πœƒ is 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ).cossin

We can see that the exponential and polar forms share the same parameters π‘Ÿ and πœƒ, which makes the conversion simpler.

From the exponential form 30𝑒οŽͺοŽ₯, we obtain modulus 30 and argument βˆ’πœ‹6. Hence, the polar form of 𝑧 is 𝑧=30ο€»ο€»βˆ’πœ‹6+π‘–ο€»βˆ’πœ‹6.cossin

To convert the polar form of a complex number to the Cartesian form, we need to multiply through the parentheses and evaluate the trigonometric ratios. Using the unit circle for trigonometric ratios, we can find the trigonometric ratios cossinο€»βˆ’πœ‹6=√32,ο€»βˆ’πœ‹6=βˆ’12.

Substituting these values into the polar form of 𝑧 and simplifying, we obtain 𝑧=30ο€Ώβˆš32βˆ’π‘–12=30√32βˆ’π‘–302=15√3βˆ’15𝑖.

Hence, the Cartesian form of the given complex number is 𝑧𝑧=15√3βˆ’15𝑖.

In the previous example, we multiplied complex numbers in the exponential form. Let us now turn our attention to the division, or the quotient, of a pair of nonzero complex numbers in the exponential form. We recall the properties of the modulus and argument under division, |||𝑧𝑧|||=|𝑧||𝑧|,𝑧𝑧=(𝑧)βˆ’(𝑧).argargarg

This leads to the division rule for complex numbers in the exponential form.

Rule: Division of Complex Numbers in Exponential Form

Let 𝑧=π‘Ÿπ‘’οŠ§οŠ§οƒοΌοŽ  and 𝑧=π‘Ÿπ‘’οŠ¨οŠ¨οƒοΌοŽ‘ be nonzero complex numbers in the exponential form. The quotient π‘§π‘§οŠ§οŠ¨ in the exponential form is π‘Ÿπ‘’π‘Ÿπ‘’=π‘Ÿπ‘Ÿπ‘’.οŠ§οƒοΌοŠ¨οƒοΌοŠ§οŠ¨οƒ()

Similar to the multiplication rule for complex numbers in the exponential form, this rule can be seen as a generalization of the law of exponents that states π‘Žπ‘Ž=π‘Ž,π‘Ž>0.ο—ο˜ο—οŠ±ο˜forany

This rule also is valid when the exponent is complex valued, as long as the base of the exponent is a positive real number.

In the next example, we will consider a quotient of complex numbers in the Cartesian form. We will first convert the complex number into the exponential form and carry out the division using this rule.

Example 6: Division of Complex Numbers

Given that 𝑍=√2𝑖1βˆ’π‘–, write 𝑍 in exponential form.

Answer

We have two methods for this example: 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 know that the division of complex numbers is simpler in the exponential form, we will choose the first method. We recall the division rule for a pair of nonzero complex numbers in the exponential form: π‘Ÿπ‘’π‘Ÿπ‘’=π‘Ÿπ‘Ÿπ‘’.οŠ§οƒοΌοŠ¨οƒοΌοŠ§οŠ¨οƒ()

We begin by converting the complex numbers in the numerator and the denominator of the given quotient. First, consider the complex number in the numerator, π‘–βˆš2. This is a purely imaginary number, where the imaginary part is positive. In an Argand diagram, this number is plotted in the positive imaginary axis, which means that its argument is πœ‹2. Also, since it is purely imaginary, the absolute value of the imaginary part of this number is equal to the modulus. This means that the modulus is √2. Hence, we can express the numerator in the exponential form as π‘–βˆš2=√2𝑒.ο‘½οΌοŽ‘

Next, let us convert the denominator of the quotient to the exponential form. Recall that the modulus of a complex number π‘Ž+𝑏𝑖 is βˆšπ‘Ž+π‘οŠ¨οŠ¨. For the denominator 1βˆ’π‘–, we have π‘Ž=1 and 𝑏=βˆ’1; thus, its modulus is 1+(βˆ’1)=√2.

For the argument of this number, we first note that 1βˆ’π‘– lies in the fourth quadrant in an Argand diagram since it has a positive real part and a negative imaginary part. Recall that the argument of a complex number π‘Ž+𝑏𝑖 in the fourth quadrant is given by arctanπ‘π‘Ž. Thus, argarctan(1βˆ’π‘–)=ο€Όβˆ’11=βˆ’πœ‹4.

Hence, we can express the denominator in the exponential form as 1βˆ’π‘–=√2𝑒.οŽͺο‘½οΌοŽ£

Now that we have obtained the exponential form of both complex numbers, we can write 𝑍=π‘–βˆš21βˆ’π‘–=√2π‘’βˆš2𝑒.ο‘½οΌοŽ‘οŽͺο‘½οΌοŽ£

Therefore, we can now apply the division rule with the values π‘Ÿ=√2,π‘Ÿ=√2,πœƒ=πœ‹2,πœƒ=βˆ’πœ‹4.

Substituting these into the division rule above, we obtain 𝑍=√2√2𝑒=𝑒.ο‘½οΌοŽ‘οŽͺο‘½οΌοŽ£οŽ’ο‘½οΌοŽ£οŠ±()

We recall that the standard range of arguments is ]βˆ’πœ‹,πœ‹], so our argument 3πœ‹4 is within the correct range.

Hence, the exponential form of the given quotient is π‘’οŽ’ο‘½οΌοŽ£.

We have considered the properties of the exponential form under the multiplication and the division of complex numbers. Let us now consider the property of the exponential form relating to the complex conjugate.

We recall the properties of the modulus and argument of a complex number under the conjugate operation. Given any nonzero complex number 𝑧, we have ||𝑧||=|𝑧|,𝑧=βˆ’π‘§.argarg

These properties lead to the following rule.

Rule: Conjugate of Complex Numbers in Exponential Form

Given a nonzero complex number in the polar form 𝑧=π‘Ÿπ‘’οƒοΌ, its conjugate 𝑧 in the polar form is π‘Ÿπ‘’=π‘Ÿπ‘’.οƒοΌοŠ±οƒοΌ

This statement tells us that the complex conjugate operation can go through the exponential function. We can see that the right-hand side of the statement can be obtained by taking the complex conjugate of the exponent only. This also is true as long as the base of the exponent is a positive real number.

Let us consider an example involving the conjugate property of the exponential form.

Example 7: Adding Complex Numbers in Exponential Form

Find the numerical value of 𝑒+π‘’οŽ οŽ ο‘½οŽ₯οŽͺοŽ οŽ ο‘½οŽ₯

Answer

In this example, we are adding two complex numbers in the exponential form. We can notice that the modulus of both complex numbers is the same, and their arguments have opposite signs. We recall the conjugate rule of complex numbers in the exponential form: π‘Ÿπ‘’=π‘Ÿπ‘’.οƒοΌοŠ±οƒοΌ

Using this rule, we can notice that the second number in the sum is the conjugate of the first number in the sum. We recall the property of complex conjugates: for any complex number 𝑧, 𝑧+𝑧=2(𝑧).Re

Hence, we can simplify our expression: 𝑒+𝑒=2𝑒.οŽ οŽ ο‘½οŽ₯οŽͺοŽ οŽ ο‘½οŽ₯οŽ οŽ ο‘½οŽ₯Re

We only need to compute the real part of this number. We recall Euler’s formula: 𝑒=πœƒ+π‘–πœƒ.cossin

From Euler’s formula, we know that Recos𝑒=11πœ‹6.οŽ οŽ ο‘½οŽ₯

Hence, 𝑒+𝑒=2ο€Ό11πœ‹6=2Γ—βˆš32=√3.οŽ οŽ ο‘½οŽ₯οŽͺοŽ οŽ ο‘½οŽ₯cos

In our final example, we will use various properties of the exponential form to solve a problem.

Example 8: Properties of Complex Numbers in Exponential Form

Given that 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.

Answer

In this example, we must find all possible expressions of a complex number from given information about a quotient. The quotient also includes a square, which is a multiplication of complex numbers. We know that the multiplication and division are simplified when using the exponential form, so we will solve the problem using various properties of the exponential form.

Recall that the exponential form of a nonzero complex number 𝑧 with modulus π‘Ÿ and argument πœƒ is 𝑧=π‘Ÿπ‘’.

Let us begin by converting π‘§οŠ§ into the exponential form. For this, we need to find its modulus and argumment. Recall that the modulus of a complex number π‘Ž+𝑏𝑖 is βˆšπ‘Ž+π‘οŠ¨οŠ¨. For 𝑧=βˆ’3√3βˆ’3π‘–οŠ§, we have π‘Ž=βˆ’3√3 and 𝑏=βˆ’3. Hence, |𝑧|=ο„žο€»βˆ’3√3+(βˆ’3)=√27+9=√36=6.

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 is 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.

Hence, we can write π‘§οŠ§ in the exponential form: 𝑧=6𝑒.οŠ§οƒοŽͺοŽ€ο‘½οŽ₯

Now, let us say that the complex number π‘§οŠ¨ has modulus π‘Ÿ and argument πœƒ, so that its exponential form is written as 𝑧=π‘Ÿπ‘’.οŠ¨οƒοΌ

We will compute the quotient π‘§π‘§οŠ§οŠ¨οŠ¨ to relate to the given information. We begin with the denominator π‘§οŠ¨οŠ¨ by recalling the multiplication rule for complex numbers in the exponential form, ο€Ήπ‘Ÿπ‘’ο…ο€Ήπ‘Ÿπ‘’ο…=π‘Ÿπ‘Ÿπ‘’.οŠ§οƒοΌοŠ¨οƒοΌοŠ§οŠ¨οƒ()

Since 𝑧=π‘§Γ—π‘§οŠ¨οŠ¨οŠ¨οŠ¨ and 𝑧=π‘Ÿπ‘’οŠ¨οƒοΌ, we can write 𝑧=ο€Ήπ‘Ÿπ‘’ο…ο€Ήπ‘Ÿπ‘’ο…=π‘Ÿπ‘’.οŠ¨οŠ¨οƒοΌοƒοΌοŠ¨οŠ¨οΌοƒ

Next, to compute the quotient π‘§π‘§οŠ§οŠ¨οŠ¨, we recall the division rule for a pair of nonzero complex numbers in the exponential form: π‘Ÿπ‘’π‘Ÿπ‘’=π‘Ÿπ‘Ÿπ‘’.οŠ§οƒοΌοŠ¨οƒοΌοŠ§οŠ¨οƒ()

Since we have already obtained 𝑧=6π‘’οŠ§οƒοŽͺοŽ€ο‘½οŽ₯ and 𝑧=π‘Ÿπ‘’οŠ¨οŠ¨οŠ¨οŠ¨οΌοƒ, we can write 𝑧𝑧=6π‘’π‘Ÿπ‘’=6π‘Ÿπ‘’.οŠ§οŠ¨οŠ¨οƒοŠ¨οŠ¨οΌοƒοŠ¨οƒ()οŽͺοŽ€ο‘½οŽ₯οŽͺοŽ€ο‘½οŽ₯

This gives us the quotient π‘§π‘§οŠ§οŠ¨οŠ¨ in the exponential form. In particular, this tells us that the modulus of this quotient is 6π‘ŸοŠ¨. We are given that this modulus is equal to 3|𝑧|. Since 𝑧=6π‘’οŠ§οƒοŽͺοŽ€ο‘½οŽ₯, we know |𝑧|=6. Hence, 6π‘Ÿ=3Γ—6=18.

Simplifying so that the subject of the equation is π‘Ÿ, 6=18π‘Ÿπ‘Ÿ=618=13π‘Ÿ=Β±1√3.

Since π‘Ÿ is the modulus of the complex number π‘§οŠ¨, it cannot be negative. Thus, we have π‘Ÿ=1√3.

Now, the only information we have not used is Imο€Ύπ‘§π‘§οŠ=0. This tells us that the imaginary part of the quotient is equal to zero, which means that the complex number on the Argand diagram would lie on the real axis. In order for the argument to lie on the real axis, the argument of this complex number must be an integer multiple of πœ‹. Hence, argforsomeintegerο€Ύπ‘§π‘§οŠ=βˆ’5πœ‹6βˆ’2πœƒ=π‘˜πœ‹,π‘˜.

Let us rearrange the equation so that πœƒ is the subject:

βˆ’5πœ‹6βˆ’2πœƒ=π‘˜πœ‹βˆ’2πœƒ=π‘˜πœ‹+5πœ‹6πœƒ=βˆ’π‘˜πœ‹2βˆ’5πœ‹12.(1)

Since πœƒ is the argument of the complex number π‘§οŠ¨, it should be in the range ]βˆ’πœ‹,πœ‹]. We need to find all possible values of πœƒ in this range. That is, we need to identify integer values π‘˜ satisfying βˆ’πœ‹<βˆ’π‘˜πœ‹2βˆ’5πœ‹12β‰€πœ‹.

Adding 5πœ‹12 to each part of the inequality above, βˆ’7πœ‹12<βˆ’π‘˜πœ‹2≀17πœ‹12.

We can multiply βˆ’2πœ‹ to each part of the inequality, reversing the direction of the inequalities: 76>π‘˜β‰₯βˆ’176.

Hence, the set of integer values π‘˜ satisfying above is {βˆ’2,βˆ’1,0,1}. Substituting these values of π‘˜ into (1), we obtain πœƒ=7πœ‹12,πœ‹12,βˆ’5πœ‹12,βˆ’11πœ‹12.

We recall that the standard range of arguments is ]βˆ’πœ‹,πœ‹], so all arguments of complex numbers above are within the correct range. Together with the modulus 1√3, the exponential forms of all possible values of π‘§οŠ¨ are 𝑧=1√3𝑒,1√3𝑒,1√3𝑒,1√3𝑒.οŠ¨οƒοƒοƒοƒοŽͺοŽ€ο‘½οŽ οŽ‘ο‘½οŽ οŽ‘οŽ¦ο‘½οŽ οŽ‘οŽͺοŽ οŽ ο‘½οŽ οŽ‘

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • For any real number πœƒ, Euler’s formula tells us that 𝑒=πœƒ+π‘–πœƒ.cossin The standard range of the argument in the exponential form is ]βˆ’πœ‹,πœ‹] radians.
  • A complex number 𝑧 with modulus π‘Ÿ and argument πœƒ is written in the exponential form as 𝑧=π‘Ÿπ‘’.
  • The exponential form of complex numbers simplifies multiplication, division, and conjugation by the following rules:
    • Multiplication rule: ο€Ήπ‘Ÿπ‘’ο…ο€Ήπ‘Ÿπ‘’ο…=π‘Ÿπ‘Ÿπ‘’οŠ§οƒοΌοŠ¨οƒοΌοŠ§οŠ¨οƒ()
    • Division rule: π‘Ÿπ‘’π‘Ÿπ‘’=π‘Ÿπ‘Ÿπ‘’οŠ§οƒοΌοŠ¨οƒοΌοŠ§οŠ¨οƒ()
    • Conjugation rule: π‘Ÿπ‘’=π‘Ÿπ‘’οƒοΌοŠ±οƒοΌ

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