Lesson Explainer: Operations on Complex Numbers in Polar Form Mathematics

In this explainer, we will learn how to perform calculations with complex numbers in polar form.

We recall that when we multiply a pair of complex numbers in the Cartesian form, we can simplify the resulting complex number in the Cartesian form by multiplying through the parentheses and collecting the real and imaginary parts separately. Also, when we divide a pair of complex numbers in the Cartesian form, we can make the denominator of the fraction real by multiplying the top and bottom of the fraction by the complex conjugate of the denominator. We can then multiply through the parentheses and collect the real and imaginary parts separately to express the resulting complex number in the Cartesian form.

This process is much simpler if we understand the polar form of complex numbers when we use the properties of the multiplication and division of complex numbers in relation to the modulus and the argument of complex numbers. In this explainer, we will prove the relationships between the multiplication/division of complex numbers and their arguments and moduli using polar forms. We begin by recalling the polar form of a complex number.

Definition: Polar Form of a Complex Number

A nonzero complex number 𝑧 with modulus |𝑧|=𝑟 and argument arg𝑧=𝜃 can be expressed in the polar form as 𝑧=𝑟(𝜃+𝑖𝜃).cossin

We recall that we can convert the Cartesian form of a complex number to the polar form by computing its modulus and argument. We can also convert the polar form of a complex number to the Cartesian form by multiplying through the parentheses and evaluating the trigonometric ratios.

Let us begin by demonstrating the relationship in contexts of the multiplication of complex numbers.

Theorem: Multiplication of Complex Numbers in Polar Form

Let 𝑧=𝑟(𝜃+𝑖𝜃)cossin and 𝑧=𝑟(𝜃+𝑖𝜃)cossin be nonzero complex numbers. Then, their product 𝑧𝑧 in the polar form is written as 𝑧𝑧=𝑟𝑟((𝜃+𝜃)+𝑖(𝜃+𝜃)).cossin

Let us prove this theorem. The product of 𝑧 and 𝑧 is written as 𝑧𝑧=(𝑟(𝜃+𝑖𝜃))(𝑟(𝜃+𝑖𝜃))=𝑟𝑟(𝜃+𝑖𝜃)(𝜃+𝑖𝜃).cossincossincossincossin

Multiplying through the parentheses, we obtain 𝑧𝑧=𝑟𝑟𝜃𝜃+𝑖𝜃𝜃+𝑖𝜃𝜃+𝑖𝜃𝜃.coscoscossincossinsinsin

Using 𝑖=1 and gathering the real and imaginary terms, we have

𝑧𝑧=𝑟𝑟((𝜃𝜃𝜃𝜃)+𝑖(𝜃𝜃+𝜃𝜃)).coscossinsincossincossin(1)

We state the summation formulae for the sine and cosine functions: coscoscossinsinsinsincoscossin(𝐴+𝐵)=𝐴𝐵𝐴𝐵,(𝐴+𝐵)=𝐴𝐵+𝐴𝐵.

We can apply the cosine summation formula to the real part and the sine summation formula to the imaginary part inside the parentheses of equation (1). Then, we can rewrite 𝑧𝑧=𝑟𝑟((𝜃+𝜃)+𝑖(𝜃+𝜃)).cossin

This proves the theorem.

In our first example, we will apply this theorem to multiply two complex numbers in the polar form.

Example 1: Multiplying Complex Numbers in Polar Form

Given that 𝑧=2𝜋6+𝑖𝜋6cossin and 𝑧=13𝜋3+𝑖𝜋3cossin, find 𝑧𝑧.

Answer

Recall that, given a pair of nonzero complex numbers 𝑧=𝑟(𝜃+𝑖𝜃)cossin and 𝑧=𝑟(𝜃+𝑖𝜃)cossin, the product is 𝑧𝑧=𝑟𝑟((𝜃+𝜃)+𝑖(𝜃+𝜃)).cossin

We are given the polar form of the complex numbers 𝑧 and 𝑧. From the given polar form, we can obtain 𝑟=2 and 𝜃=𝜋6 for 𝑧, while 𝑟=13 and 𝜃=𝜋3. Substituting these values into the formula, we have 𝑧𝑧=23𝜋6+𝜋3+𝑖𝜋6+𝜋3.cossin

Rationalizing the denominator and summing the fractions, we have 𝑧𝑧=233𝜋2+𝑖𝜋2.cossin

In the previous example, we computed the product of two complex numbers in the polar form. We note that this process is simpler than the multiplication of complex numbers in the Cartesian form, which would involve multiplying through the parentheses and collecting the real and the imaginary terms.

Let us examine the polar form of the product 𝑧𝑧=𝑟𝑟((𝜃+𝜃)+𝑖(𝜃+𝜃)).cossin

From this polar form, we can see that the modulus of the product 𝑧𝑧 is 𝑟𝑟, which is the product of the moduli of 𝑧 and 𝑧. Also, the argument of the product 𝑧𝑧 is the sum of the arguments of 𝑧 and 𝑧. This leads to the following fact.

Fact: Relationship between Product of Complex Numbers and Their Moduli and Arguments

For any pair of nonzero complex numbers 𝑧 and 𝑧, we have |𝑧𝑧|=|𝑧||𝑧|,(𝑧𝑧)=𝑧+𝑧.argargarg

In the next example, we will apply this fact to find the modulus of the product of two complex numbers.

Example 2: Multiplying Complex Numbers in Polar Form

If 𝑍=7(𝜃+𝑖𝜃)cossin, 𝑍=16(𝜃+𝑖𝜃)cossin, and 𝜃+𝜃=𝜋, then what is 𝑍𝑍?

Answer

Recall that, for any pair of nonzero complex numbers 𝑍 and 𝑍, |𝑍𝑍|=|𝑍||𝑍|,(𝑍𝑍)=𝑍+𝑍.argargarg

In this example, we are given the complex numbers 𝑍 and 𝑍 in the polar form. We recall that a nonzero complex number 𝑧 has the polar form 𝑧=|𝑧|((𝑧)+𝑖(𝑧)).cosargsinarg

From the given polar form, we can see that |𝑍|=7 and |𝑍|=16, which means |𝑍𝑍|=|𝑍||𝑍|=7×16=112.

We can also see that arg𝑍=𝜃 and arg𝑍=𝜃. So, arg𝑍𝑍=𝜃+𝜃.

Since we are given that 𝜃+𝜃=𝜋, we have arg𝑍𝑍=𝜋. This leads to the polar form of the product 𝑍𝑍=112(𝜋+𝑖𝜋).cossin

We know that cos𝜋=1 and sin𝜋=0. Substituting these values into the equation above, we have 𝑍𝑍=112(1+𝑖0)=112.

Hence, 𝑍𝑍 is the real number 112.

In previous examples, we found the products of complex numbers in the polar form. We now turn our attention to the division and quotient of complex numbers in the polar form.

Definition: The Quotient of Complex Numbers in Polar Form

Given a pair of nonzero complex numbers in the polar form 𝑧=𝑟(𝜃+𝑖𝜃)cossin and 𝑧=𝑟(𝜃+𝑖𝜃)cossin, their quotient can be written in the polar form as 𝑧𝑧=𝑟𝑟((𝜃𝜃)+𝑖(𝜃𝜃)).cossin

Let us prove this theorem. We begin with the quotient of 𝑧 and 𝑧 written as 𝑧𝑧=𝑟(𝜃+𝑖𝜃)𝑟(𝜃+𝑖𝜃)=𝑟𝑟𝜃+𝑖𝜃𝜃+𝑖𝜃.cossincossincossincossin

To make the denominator of this fraction real, we multiply the numerator and denominator by the conjugate of the denominator to obtain 𝑧𝑧=𝑟𝑟(𝜃+𝑖𝜃)(𝜃𝑖𝜃)(𝜃+𝑖𝜃)(𝜃𝑖𝜃).cossincossincossincossin

Multiplying through the parentheses, we obtain 𝑧𝑧=𝑟𝑟𝜃𝜃𝑖𝜃𝜃+𝑖𝜃𝜃𝑖𝜃𝜃𝜃+𝑖𝜃𝜃𝑖𝜃𝜃𝑖𝜃.coscoscossinsincossinsincossincossincossin

Using 𝑖=1 and gathering the real and imaginary terms, we have 𝑧𝑧=𝑟𝑟(𝜃𝜃+𝜃𝜃)+𝑖(𝜃𝜃𝜃𝜃)𝜃+𝜃.coscossinsinsincoscossincossin

Using the trigonometric identity sincos𝜃+𝜃=1, we can simplify this expression to

𝑧𝑧=𝑟𝑟((𝜃𝜃+𝜃𝜃)+𝑖(𝜃𝜃𝜃𝜃)).coscossinsinsincoscossin(2)

Finally, we state the difference identities for sine and cosine: coscoscossinsinsinsincoscossin(𝐴𝐵)=𝐴𝐵+𝐴𝐵,(𝐴𝐵)=𝐴𝐵𝐴𝐵.

We apply the cosine difference identity in the real part and the sine difference identity in the imaginary part of the complex number within the parentheses on the right-hand side of equation (2). Then, we can rewrite: 𝑧𝑧=𝑟𝑟((𝜃𝜃)+𝑖(𝜃𝜃)).cossin

This proves the theorem.

Let us consider an example where we will find the quotient of two complex numbers in the polar form using this method.

Example 3: Finding the Quotient of Two Complex Numbers in Polar Form

Given that 𝑧=20𝜋2+𝑖𝜋2cossin and 𝑧=4𝜋6+𝑖𝜋6cossin, find 𝑧𝑧 in polar form.

Answer

Recall that, given a pair of nonzero complex numbers in the polar form 𝑧=𝑟(𝜃+𝑖𝜃)cossin and 𝑧=𝑟(𝜃+𝑖𝜃)cossin, the quotient can be written in the polar form as 𝑧𝑧=𝑟𝑟((𝜃𝜃)+𝑖(𝜃𝜃)).cossin

In this example, we are given the power form for the complex numbers 𝑧 and 𝑧. From the given polar form, we can identify 𝑟=20 and 𝜃=𝜋2 for 𝑧, while 𝑟=4 and 𝜃=𝜋6. Substituting these values into the formula for the polar form of the quotient, we have 𝑧𝑧=204𝜋2𝜋6+𝑖𝜋2𝜋6.cossin

Simplifying, we have 𝑧𝑧=5𝜋3+𝑖𝜋3.cossin

In the previous example, we computed the quotient of two complex numbers in the polar form. We note that this process is simpler than the division of complex numbers in the Cartesian form, which would involve multiplying the numerator and denominator by the conjugate of the denominator and then multiplying through the parentheses. Using this method, we can see that the division of complex numbers is much simpler in the polar form.

Let us examine the polar form of the quotient: 𝑧𝑧=𝑟𝑟((𝜃𝜃)+𝑖(𝜃𝜃)).cossin

From this polar form, we can see that the modulus of the quotient 𝑧𝑧 is 𝑟𝑟, which is the quotient of the moduli of 𝑧 and 𝑧. Also, the argument of the quotient 𝑧𝑧 is the difference of the arguments of 𝑧 and 𝑧. This leads to the following fact.

Fact: Relationship between Quotient of Complex Numbers and Their Moduli and Arguments

For any pair of nonzero complex numbers 𝑧 and 𝑧, we have |||𝑧𝑧|||=|𝑧||𝑧|,𝑧𝑧=𝑧𝑧.argargarg

In the next example, we will use these facts to find the polar form of a quotient of two complex numbers.

Example 4: Dividing Complex Numbers in Polar Form and Finding Their Quotient in Cartesian Form

Given that 𝑍=5(5𝜃+𝑖5𝜃)cossin, 𝑍=4𝜃+𝑖4𝜃cossin, tan𝜃=43, and 𝜃0,𝜋2, find 𝑍𝑍.

  1. 4+3𝑖
  2. 3+4𝑖
  3. 35+45𝑖
  4. 45+35𝑖

Answer

Recall that, given any pair of nonzero complex numbers 𝑧 and 𝑧, |||𝑧𝑧|||=|𝑧||𝑧|,𝑧𝑧=𝑧𝑧.argargarg

In this example, the complex numbers 𝑍 and 𝑍 are given in the polar form. We recall that a nonzero complex number 𝑧 has the polar form 𝑧=|𝑧|((𝑧)+𝑖(𝑧)).cosargsinarg

From the given polar form, we can find the moduli |𝑍|=5, |𝑍|=1. Hence, |||𝑧𝑧|||=51=5.

Also from the given polar form, we can find the arguments arg(𝑍)=5𝜃, arg(𝑍)=4𝜃. Therefore, argargarg𝑧𝑧=(𝑍)(𝑍)=5𝜃4𝜃=𝜃.

Hence, the polar form of the quotient 𝑍𝑍 is 𝑍𝑍=5(𝜃+𝑖𝜃).cossin

To finish the problem, we need to find the trigonometric ratios cos𝜃 and sin𝜃 from the provided information about the tangent function. We are given that 𝜃0,𝜋2. Since the tangent function is defined at 𝜃, while the tangent function is not defined at 𝜋2, we know that 𝜃𝜋2. This means that 𝜃0,𝜋2. In other words, 𝜃 is an acute angle. For an acute angle, we can relate trigonometric ratios to right triangle trigonometry. Recall the trigonometric ratios for an acute angle 𝜃: sinoppositehypotenusecosadjacenthypotenusetanoppositeadjacent𝜃=,𝜃=,𝜃=.

We are given that tan𝜃=43, so we can draw the right triangle with an angle theta whose opposite side is of length 4 and whose adjacent side is of length 3. Using the Pythagorean theorem, the length of the hypotenuse must be 3+4=25=5.

Using this triangle, we find sinoppositehypotenusecosadjacenthypotenuse𝜃==45,𝜃==35.

Substituting these values into the polar form for the quotient, we have 𝑍𝑍=535+45𝑖=3+4𝑖.

This leads to option B.

In the previous two examples, we found the quotients of complex numbers using the polar form. We can also use the rules of division to find a general form for the reciprocal of a complex number as the next example will show.

Example 5: The Reciprocal of a Complex Number in Polar Form

Given that 𝑧=7𝜋6+𝑖7𝜋6cossin, find 1𝑧.

Answer

Recall that, given a pair of nonzero complex numbers in the polar form 𝑧=𝑟(𝜃+𝑖𝜃)cossin and 𝑧=𝑟(𝜃+𝑖𝜃)cossin, the quotient can be written in the polar form as 𝑧𝑧=𝑟𝑟((𝜃𝜃)+𝑖(𝜃𝜃)).cossin

In this example, we need to find the reciprocal 1𝑧. Note that the reciprocal is also a fraction where 𝑧=1 and 𝑧=𝑧. In this case, 𝑧=1 is the real number, which means that it has modulus 1 and argument 0. In other words, 1 can be expressed in the polar form as 1=1(0+𝑖0).cossin

This leads to 𝑟=1 and 𝜃=0. On the other hand, the denominator 𝑧=𝑧 is given in the polar form; hence, we can obtain 𝑟=1 and 𝜃=7𝜋6. Substituting these values into the equation for the polar form of the quotient, we have 1𝑧=1107𝜋6+𝑖07𝜋6=7𝜋6+𝑖7𝜋6.cossincossin

While this is a correct answer, we also recall that, by convention, the argument of a complex number should lie in the range (𝜋,𝜋] in radians. Such an argument is called the principal argument. The argument given in the polar form above, 7𝜋6, does not lie in the range (𝜋,𝜋], so we need to add or subtract a multiple of the full revolution 2𝜋. Since the given argument is below the lower bound 𝜋, we add 2𝜋 to obtain an equivalent argument: 7𝜋6+2𝜋=7𝜋6+12𝜋6=5𝜋6.

This argument lies in the range (𝜋,𝜋], which makes it the principal argument. Using the principal argument, the polar form of the reciprocal is 1𝑧=5𝜋6+𝑖5𝜋6.cossin

In the previous example, we found the reciprocal of a complex number in the polar form using the formula for the quotient using the polar form. Applying an analogous method, we can find a general formula for the polar form of the reciprocal of a complex number.

Definition: Polar Form of a Reciprocal of a Complex Number

Given a nonzero complex number in the polar form 𝑧=𝑟(𝜃+𝑖𝜃)cossin, the reciprocal can be written in the polar form as 1𝑧=1𝑟((𝜃)+𝑖(𝜃)).cossin

The final example will demonstrate how we can use the formula for the product of complex numbers in polar form to find formulae for the powers of complex numbers.

Example 6: Using the Modulus and Argument to Calculate Powers of Complex Numbers in Cartesian Form

Consider the complex number 𝑧=1+𝑖3.

  1. Find the modulus of 𝑧.
  2. Find the argument of 𝑧.
  3. Hence, use the properties of multiplication of complex numbers in polar form to find the modulus and argument of 𝑧.
  4. Hence, find the value of 𝑧.

Answer

Part 1

Recall that the modulus of a complex number 𝑧=𝑎+𝑏𝑖 in the Cartesian form is |𝑧|=𝑎+𝑏.

In our example, 𝑎=1 and 𝑏=3, so we obtain |𝑧|=1+3=4=2.

Hence, the modulus of 1+𝑖3 is 2.

Part 2

To calculate the argument, we first consider which quadrant of an Argand diagram the complex number lies in. Since both its real and imaginary parts are positive, the complex number 1+𝑖3 lies in the first quadrant of an Argand diagram. We recall that the argument of a complex number 𝑎+𝑏𝑖 in the first quadrant is given by arctan𝑏𝑎. Then, argarctanarctanradians(𝑧)=𝑏𝑎=3=𝜋3.

Hence, the argument of 1+𝑖3 is 𝜋3.

Part 3

We recall the properties of the multiplication of complex numbers in relation to the moduli and the arguments of the complex numbers: for any pair of nonzero complex numbers 𝑧 and 𝑧, we have |𝑧𝑧|=|𝑧||𝑧|,(𝑧𝑧)=𝑧+𝑧.argargarg

In this example, we need to compute 𝑧, which can be obtained by multiplying 𝑧 three times: 𝑧=𝑧×𝑧×𝑧.

Since we can obtain the modulus of a product by taking the product of the moduli of the complex numbers, we have that ||𝑧||=|𝑧×𝑧×𝑧|=|𝑧|×|𝑧|×|𝑧|=|𝑧|.

In part 1, we obtained the modulus |𝑧|=2, so ||𝑧||=2=8.

Hence, the modulus of 𝑧 is 8.

Similarly, we know that the argument of the product of a complex number is the sum of the arguments of each complex number. So, argargargargarg𝑧=𝑧+𝑧+𝑧=3𝑧.

In part 2, we obtained that arg𝑧=𝜋3, so arg𝑧=3×𝜋3=𝜋.

Hence, the argument of 𝑧 is 𝜋.

Part 4

We recall that a nonzero complex number 𝑤 with modulus 𝑟 and argument 𝜃 has the polar form 𝑤=𝑟(𝜃+𝑖(𝜃)).cossin

In the previous two parts, we have calculated that the modulus of 𝑧 is 8, so 𝑟=8. We also obtained that the argument of 𝑧 is 𝜋; hence, 𝜃=𝜋. Substituting these values into the polar form, 𝑧=8(𝜋+𝑖𝜋).cossin

Since cos𝜋=1 and sin𝜋=0, we have 𝑧=8(1+𝑖0)=8.

Hence, 𝑧=8.

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

Key Points

  • Multiplication and division of complex numbers are often simpler when we work with complex numbers in polar form.
  • Given a pair of nonzero complex numbers in the polar form 𝑧=𝑟(𝜃+𝑖𝜃)cossin and 𝑧=𝑟(𝜃+𝑖𝜃)cossin,
    • the product of the complex numbers in the polar form is 𝑧𝑧=𝑟𝑟((𝜃+𝜃)+𝑖(𝜃+𝜃)),cossin
    • the quotient of the complex numbers in the polar form is 𝑧𝑧=𝑟𝑟((𝜃𝜃)+𝑖(𝜃𝜃)).cossin
  • For any pair of nonzero complex numbers 𝑧 and 𝑧, we have |𝑧𝑧|=|𝑧||𝑧|,(𝑧𝑧)=𝑧+𝑧,|||𝑧𝑧|||=|𝑧||𝑧|,𝑧𝑧=𝑧𝑧.argargargargargarg
  • Given a nonzero complex number in the polar form 𝑧=𝑟(𝜃+𝑖𝜃)cossin, the reciprocal can be written in the polar form as 1𝑧=1𝑟((𝜃)+𝑖(𝜃)).cossin

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