Polar Form of Complex Numbers

Learning Objectives

In this section, you will:

• Plot complex numbers in the complex plane.
• Find the absolute value of a complex number.
• Write complex numbers in polar form.
• Convert a complex number from polar to rectangular form.
• Find products of complex numbers in polar form.
• Find quotients of complex numbers in polar form.
• Find powers of complex numbers in polar form.
• Find roots of complex numbers in polar form.

“God made the integers; all else is the work of man.” This rather famous quote by nineteenth-century German mathematician Leopold Kronecker sets the stage for this section on the polar form of a complex number. Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as Pythagoras, Descartes, De Moivre, Euler, Gauss, and others. Complex numbers answered questions that for centuries had puzzled the greatest minds in science.

We first encountered complex numbers in Complex Numbers. In this section, we will focus on the mechanics of working with complex numbers: translation of complex numbers from polar form to rectangular form and vice versa, interpretation of complex numbers in the scheme of applications, and application of De Moivre’s Theorem.

Plotting Complex Numbers in the Complex Plane

Plotting a complex number a+bi𝑎+𝑏𝑖 is similar to plotting a real number, except that the horizontal axis represents the real part of the number, a,𝑎, and the vertical axis represents the imaginary part of the number, bi.𝑏𝑖.

HOW TO

Given a complex number a+bi,𝑎+𝑏𝑖, plot it in the complex plane.

1. Label the horizontal axis as the real axis and the vertical axis as the imaginary axis.
2. Plot the point in the complex plane by moving a𝑎 units in the horizontal direction and b𝑏 units in the vertical direction.

EXAMPLE 1

Plotting a Complex Number in the Complex Plane

Plot the complex number 2−3i2−3𝑖 in the complex plane.

Solution

From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction. See Figure 1.

Figure 1

TRY IT #1

Plot the point 1+5i1+5𝑖 in the complex plane.

Finding the Absolute Value of a Complex Number

The first step toward working with a complex number in polar form is to find the absolute value. The absolute value of a complex number is the same as its magnitude, or |z|.|𝑧|. It measures the distance from the origin to a point in the plane. For example, the graph of z=2+4i,𝑧=2+4𝑖, in Figure 2, shows |z|.|𝑧|.

Figure 2

ABSOLUTE VALUE OF A COMPLEX NUMBER

Given z=x+yi,𝑧=𝑥+𝑦𝑖, a complex number, the absolute value of z𝑧 is defined as

|z|=x2+y2−−−−−−√|𝑧|=𝑥2+𝑦2

It is the distance from the origin to the point (x,y).(𝑥,𝑦).

Notice that the absolute value of a real number gives the distance of the number from 0, while the absolute value of a complex number gives the distance of the number from the origin, (0,0).(0,0).

EXAMPLE 2

Finding the Absolute Value of a Complex Number with a Radical

Find the absolute value of z=5–√−i.𝑧=5−𝑖.

Solution

Using the formula, we have

|z|=x2+y2−−−−−−√|z|=(5–√)2+(−1)2−−−−−−−−−−−−√|z|=5+1−−−−√|z|=6–√|𝑧|=𝑥2+𝑦2|𝑧|=(5)2+(−1)2|𝑧|=5+1|𝑧|=6

See Figure 3.

Figure 3

TRY IT #2

Find the absolute value of the complex number z=12−5i.𝑧=12−5𝑖.

EXAMPLE 3

Finding the Absolute Value of a Complex Number

Given z=3−4i,𝑧=3−4𝑖, find |z|.|𝑧|.

Solution

Using the formula, we have

|z|=x2+y2−−−−−−√|z|=(3)2+(−4)2−−−−−−−−−−√|z|=9+16−−−−−√|z|=25−−√|z|=5|𝑧|=𝑥2+𝑦2|𝑧|=(3)2+(−4)2|𝑧|=9+16|𝑧|=25|𝑧|=5

The absolute value z𝑧 is 5. See Figure 4.

Figure 4

TRY IT #3

Given z=1−7i,𝑧=1−7𝑖, find |z|.|𝑧|.

Writing Complex Numbers in Polar Form

The polar form of a complex number expresses a number in terms of an angle θ𝜃 and its distance from the origin r.𝑟. Given a complex number in rectangular form expressed as z=x+yi,𝑧=𝑥+𝑦𝑖, we use the same conversion formulas as we do to write the number in trigonometric form:

x=rcosθy=rsinθr=x2+y2−−−−−−√𝑥=𝑟cos𝜃𝑦=𝑟sin𝜃𝑟=𝑥2+𝑦2

We review these relationships in Figure 5.

Figure 5

We use the term modulus to represent the absolute value of a complex number, or the distance from the origin to the point (x,y).(𝑥,𝑦). The modulus, then, is the same as r,𝑟, the radius in polar form. We use θ𝜃 to indicate the angle of direction (just as with polar coordinates). Substituting, we have

z=x+yiz=rcosθ+(rsinθ)iz=r(cosθ+isinθ)𝑧=𝑥+𝑦𝑖𝑧=𝑟cos𝜃+(𝑟sin𝜃)𝑖𝑧=𝑟(cos𝜃+𝑖sin𝜃)

POLAR FORM OF A COMPLEX NUMBER

Writing a complex number in polar form involves the following conversion formulas:

x=rcosθy=rsinθr=x2+y2−−−−−−√𝑥=𝑟cos𝜃𝑦=𝑟sin𝜃𝑟=𝑥2+𝑦2

Making a direct substitution, we have

z=x+yiz=(rcosθ)+i(rsinθ)z=r(cosθ+isinθ)𝑧=𝑥+𝑦𝑖𝑧=(𝑟cos𝜃)+𝑖(𝑟sin𝜃)𝑧=𝑟(cos𝜃+𝑖sin𝜃)

where r𝑟 is the modulus and θ𝜃 is the argument. We often use the abbreviation rcisθ𝑟cis𝜃 to represent r(cosθ+isinθ).𝑟(cos𝜃+𝑖sin𝜃).

EXAMPLE 4

Expressing a Complex Number Using Polar Coordinates

Express the complex number 4i4𝑖 using polar coordinates.

Solution

On the complex plane, the number z=4i𝑧=4𝑖 is the same as z=0+4i.𝑧=0+4𝑖. Writing it in polar form, we have to calculate r𝑟 first.

r=x2+y2−−−−−−√r=02+42−−−−−−√r=16−−√r=4𝑟=𝑥2+𝑦2𝑟=02+42𝑟=16𝑟=4

Next, we look at x.𝑥. If x=rcosθ,𝑥=𝑟cos𝜃, and x=0,𝑥=0, then θ=π2.𝜃=𝜋2. In polar coordinates, the complex number z=0+4i𝑧=0+4𝑖 can be written as z=4(cos(π2)+isin(π2))𝑧=4(cos(𝜋2)+𝑖sin(𝜋2)) or 4cis(π2).4cis(𝜋2). See Figure 6.

Figure 6

TRY IT #4

Express z=3i𝑧=3𝑖 as rcisθ𝑟cis𝜃 in polar form.

EXAMPLE 5

Finding the Polar Form of a Complex Number

Find the polar form of −4+4i.−4+4𝑖.

Solution

First, find the value of r.𝑟.

r=x2+y2−−−−−−√r=(−4)2+(42)−−−−−−−−−−√r=32−−√r=42–√𝑟=𝑥2+𝑦2𝑟=(−4)2+(42)𝑟=32𝑟=42

Find the angle θ𝜃 using the formula:

cosθ=xrcosθ=−442√cosθ=−12√θ=cos−1(−12√)=3π4cos𝜃=𝑥𝑟cos𝜃=−442cos𝜃=−12𝜃=cos−1(−12)=3𝜋4

Thus, the solution is 42–√cis(3π4).42cis(3𝜋4).

TRY IT #5

Write z=3–√+i𝑧=3+𝑖 in polar form.

Converting a Complex Number from Polar to Rectangular Form

Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. In other words, given z=r(cosθ+isinθ),𝑧=𝑟(cos𝜃+𝑖sin𝜃), first evaluate the trigonometric functions cosθcos𝜃 and sinθ.sin𝜃. Then, multiply through by r.𝑟.

EXAMPLE 6

Converting from Polar to Rectangular Form

Convert the polar form of the given complex number to rectangular form:

z=12(cos(π6)+isin(π6))𝑧=12(cos(𝜋6)+𝑖sin(𝜋6))

Solution

We begin by evaluating the trigonometric expressions.

cos(π6)=3–√2andsin(π6)=12cos(𝜋6)=32andsin(𝜋6)=12

After substitution, the complex number is

z=12(3–√2+12i)𝑧=12(32+12𝑖)

We apply the distributive property:

z=12(3√2+12i)  =(12)3√2+(12)12i  =63–√+6i𝑧=12(32+12𝑖)  =(12)32+(12)12𝑖  =63+6𝑖

The rectangular form of the given point in complex form is 63–√+6i.63+6𝑖.

EXAMPLE 7

Finding the Rectangular Form of a Complex Number

Find the rectangular form of the complex number given r=13𝑟=13 and tanθ=512.tan𝜃=512.

Solution

If tanθ=512,tan𝜃=512, and tanθ=yx,tan𝜃=𝑦𝑥, we first determine r=x2+y2−−−−−−√=122+52−−−−−−−√=13.𝑟=𝑥2+𝑦2=122+52=13. We then find cosθ=xrcos𝜃=𝑥𝑟 and sinθ=yr.sin𝜃=𝑦𝑟.

z=13(cosθ+isinθ)=13(1213+513i)=12+5i𝑧=13(cos𝜃+𝑖sin𝜃)=13(1213+513𝑖)=12+5𝑖

The rectangular form of the given number in complex form is 12+5i.12+5𝑖.

TRY IT #6

Convert the complex number to rectangular form:

z=4(cos11π6+isin11π6)𝑧=4(cos11𝜋6+𝑖sin11𝜋6)

Finding Products of Complex Numbers in Polar Form

Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. For the rest of this section, we will work with formulas developed by French mathematician Abraham De Moivre (1667-1754). These formulas have made working with products, quotients, powers, and roots of complex numbers much simpler than they appear. The rules are based on multiplying the moduli and adding the arguments.

PRODUCTS OF COMPLEX NUMBERS IN POLAR FORM

If z1=r1(cosθ1+isinθ1)𝑧1=𝑟1(cos𝜃1+𝑖sin𝜃1) and z2=r2(cosθ2+isinθ2),𝑧2=𝑟2(cos𝜃2+𝑖sin𝜃2), then the product of these numbers is given as:

z1z2=r1r2[cos(θ1+θ2)+isin(θ1+θ2)]z1z2=r1r2cis(θ1+θ2)𝑧1𝑧2=𝑟1𝑟2[cos(𝜃1+𝜃2)+𝑖sin(𝜃1+𝜃2)]𝑧1𝑧2=𝑟1𝑟2cis(𝜃1+𝜃2)

Notice that the product calls for multiplying the moduli and adding the angles.

EXAMPLE 8

Finding the Product of Two Complex Numbers in Polar Form

Find the product of z1z2,𝑧1𝑧2, given z1=4(cos(80°)+isin(80°))𝑧1=4(cos(80°)+𝑖sin(80°)) and z2=2(cos(145°)+isin(145°)).𝑧2=2(cos(145°)+𝑖sin(145°)).

Solution

Follow the formula

z1z2=4⋅2[cos(80°+145°)+isin(80°+145°)]z1z2=8[cos(225°)+isin(225°)]z1z2=8[cos(5π4)+isin(5π4)]z1z2=8[−2√2+i(−2√2)]z1z2=−42–√−4i2–√𝑧1𝑧2=4⋅2[cos(80°+145°)+𝑖sin(80°+145°)]𝑧1𝑧2=8[cos(225°)+𝑖sin(225°)]𝑧1𝑧2=8[cos(5𝜋4)+𝑖sin(5𝜋4)]𝑧1𝑧2=8[−22+𝑖(−22)]𝑧1𝑧2=−42−4𝑖2

Finding Quotients of Complex Numbers in Polar Form

The quotient of two complex numbers in polar form is the quotient of the two moduli and the difference of the two arguments.

QUOTIENTS OF COMPLEX NUMBERS IN POLAR FORM

If z1=r1(cosθ1+isinθ1)𝑧1=𝑟1(cos𝜃1+𝑖sin𝜃1) and z2=r2(cosθ2+isinθ2),𝑧2=𝑟2(cos𝜃2+𝑖sin𝜃2), then the quotient of these numbers is

z1z2=r1r2[cos(θ1−θ2)+isin(θ1−θ2)],z2≠0z1z2=r1r2cis(θ1−θ2),z2≠0𝑧1𝑧2=𝑟1𝑟2[cos(𝜃1−𝜃2)+𝑖sin(𝜃1−𝜃2)],𝑧2≠0𝑧1𝑧2=𝑟1𝑟2cis(𝜃1−𝜃2),𝑧2≠0

Notice that the moduli are divided, and the angles are subtracted.

HOW TO

Given two complex numbers in polar form, find the quotient.

1. Divide r1r2.𝑟1𝑟2.
2. Find θ1−θ2.𝜃1−𝜃2.
3. Substitute the results into the formula: z=r(cosθ+isinθ).𝑧=𝑟(cos𝜃+𝑖sin𝜃). Replace r𝑟 with r1r2,𝑟1𝑟2, and replace θ𝜃 with θ1−θ2.𝜃1−𝜃2.
4. Calculate the new trigonometric expressions and multiply through by r.𝑟.

EXAMPLE 9

Finding the Quotient of Two Complex Numbers

Find the quotient of z1=2(cos(213°)+isin(213°))𝑧1=2(cos(213°)+𝑖sin(213°)) and z2=4(cos(33°)+isin(33°)).𝑧2=4(cos(33°)+𝑖sin(33°)).

Solution

Using the formula, we have

z1z2=24[cos(213°−33°)+isin(213°−33°)]z1z2=12[cos(180°)+isin(180°)]z1z2=12[−1+0i]z1z2=−12+0iz1z2=−12𝑧1𝑧2=24[cos(213°−33°)+𝑖sin(213°−33°)]𝑧1𝑧2=12[cos(180°)+𝑖sin(180°)]𝑧1𝑧2=12[−1+0𝑖]𝑧1𝑧2=−12+0𝑖𝑧1𝑧2=−12

TRY IT #7

Find the product and the quotient of z1=23–√(cos(150°)+isin(150°))𝑧1=23(cos(150°)+𝑖sin(150°)) and z2=2(cos(30°)+isin(30°)).𝑧2=2(cos(30°)+𝑖sin(30°)).

Finding Powers of Complex Numbers in Polar Form

Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem. It states that, for a positive integer n,zn𝑛,𝑧𝑛 is found by raising the modulus to the nth𝑛th power and multiplying the argument by n.𝑛. It is the standard method used in modern mathematics.

DE MOIVRE’S THEOREM

If z=r(cosθ+isinθ)𝑧=𝑟(cos𝜃+𝑖sin𝜃) is a complex number, then

zn=rn[cos(nθ)+isin(nθ)]zn=rncis(nθ)𝑧𝑛=𝑟𝑛[cos(𝑛𝜃)+𝑖sin(𝑛𝜃)]𝑧𝑛=𝑟𝑛cis(𝑛𝜃)

where n𝑛 is a positive integer.

EXAMPLE 10

Evaluating an Expression Using De Moivre’s Theorem

Evaluate the expression (1+i)5(1+𝑖)5 using De Moivre’s Theorem.

Solution

Since De Moivre’s Theorem applies to complex numbers written in polar form, we must first write (1+i)(1+𝑖) in polar form. Let us find r.𝑟.

r=x2+y2−−−−−−√r=(1)2+(1)2−−−−−−−−−√r=2–√𝑟=𝑥2+𝑦2𝑟=(1)2+(1)2𝑟=2

Then we find θ.𝜃. Using the formula tanθ=yxtan𝜃=𝑦𝑥 gives

tanθ=11tanθ=1θ=π4tan𝜃=11tan𝜃=1𝜃=𝜋4

Use De Moivre’s Theorem to evaluate the expression.

(a+bi)n=rn[cos(nθ)+isin(nθ)](1+i)5=(2–√)5[cos(5⋅π4)+isin(5⋅π4)](1+i)5=42–√[cos(5π4)+isin(5π4)](1+i)5=42–√[−2√2+i(−2√2)](1+i)5=−4−4i(𝑎+𝑏𝑖)𝑛=𝑟𝑛[cos(𝑛𝜃)+𝑖sin(𝑛𝜃)](1+𝑖)5=(2)5[cos(5⋅𝜋4)+𝑖sin(5⋅𝜋4)](1+𝑖)5=42[cos(5𝜋4)+𝑖sin(5𝜋4)](1+𝑖)5=42[−22+𝑖(−22)](1+𝑖)5=−4−4𝑖

Finding Roots of Complex Numbers in Polar Form

To find the nth root of a complex number in polar form, we use the nth𝑛th Root Theorem or De Moivre’s Theorem and raise the complex number to a power with a rational exponent. There are several ways to represent a formula for finding nth𝑛th roots of complex numbers in polar form.

THE NTH ROOT THEOREM

To find the nth𝑛th root of a complex number in polar form, use the formula given as

z1n=r1n[cos(θn+2kπn)+isin(θn+2kπn)]𝑧1𝑛=𝑟1𝑛[cos(𝜃𝑛+2𝑘𝜋𝑛)+𝑖sin(𝜃𝑛+2𝑘𝜋𝑛)]

where k=0,1,2,3,…,n−1.𝑘=0,1,2,3,…,𝑛−1. We add 2kπn2𝑘𝜋𝑛 to θn𝜃𝑛 in order to obtain the periodic roots.

EXAMPLE 11

Finding the nth Root of a Complex Number

Evaluate the cube roots of z=8(cos(2π3)+isin(2π3)).𝑧=8(cos(2𝜋3)+𝑖sin(2𝜋3)).

Solution

We have

z13=813[cos(2π33+2kπ3)+isin(2π33+2kπ3)]z13=2[cos(2π9+2kπ3)+isin(2π9+2kπ3)]𝑧13=813[cos(2𝜋33+2𝑘𝜋3)+𝑖sin(2𝜋33+2𝑘𝜋3)]𝑧13=2[cos(2𝜋9+2𝑘𝜋3)+𝑖sin(2𝜋9+2𝑘𝜋3)]

There will be three roots: k=0,1,2.𝑘=0,1,2. When k=0,𝑘=0, we have

z13=2(cos(2π9)+isin(2π9))𝑧13=2(cos(2𝜋9)+𝑖sin(2𝜋9))

When k=1,𝑘=1, we have

z13=2[cos(2π9+6π9)+isin(2π9+6π9)]    Add 2(1)π3 to each angle.z13=2(cos(8π9)+isin(8π9))𝑧13=2[cos(2𝜋9+6𝜋9)+𝑖sin(2𝜋9+6𝜋9)]    Add 2(1)𝜋3 to each angle.𝑧13=2(cos(8𝜋9)+𝑖sin(8𝜋9))

When k=2,𝑘=2, we have

z13=2[cos(2π9+12π9)+isin(2π9+12π9)]z13=2(cos(14π9)+isin(14π9))Add 2(2)π3 to each angle.𝑧13=2[cos(2𝜋9+12𝜋9)+𝑖sin(2𝜋9+12𝜋9)]Add 2(2)𝜋3 to each angle.𝑧13=2(cos(14𝜋9)+𝑖sin(14𝜋9))

Remember to find the common denominator to simplify fractions in situations like this one. For k=1,𝑘=1, the angle simplification is

2π33+2(1)π3=2π3(13)+2(1)π3(33)=2π9+6π9=8π92𝜋33+2(1)𝜋3=2𝜋3(13)+2(1)𝜋3(33)=2𝜋9+6𝜋9=8𝜋9

TRY IT #8

Find the four fourth roots of 16(cos(120°)+isin(120°)).