Explainer: Polar Form of Complex Numbers

In this explainer, we will learn how to represent a complex number in polar form, calculate the modulus and argument, and use this to change the form of a complex number.

We can represent a complex number such as 𝑧=4+4𝑖 (where 𝑖 is the square root of negative one) on an Argand diagram as shown below.

We can refer to points in the plane in both Cartesian and polar forms. In a similar way, complex numbers can be written in both algebraic and polar forms.

Recall, when changing the coordinates of a point (π‘₯,𝑦) to polar form (π‘Ÿ,πœƒ), that we calculate π‘Ÿ using the Pythagorean theorem as follows: π‘Ÿ=√π‘₯+𝑦, and we calculate πœƒ using the inverse tangent function πœƒ=𝑦π‘₯.arctan

Applying the same method to the point 𝑧=4+4𝑖, we calculate π‘Ÿ=√4+4=√32=4√2, and we calculate πœƒ=ο€Ό44=πœ‹4.arctan

For complex numbers, we have special names for π‘Ÿ and πœƒ: we refer to π‘Ÿ as the modulus of the complex number (which we write as |𝑧|=π‘Ÿ) and πœƒ as the argument (which we write as arg𝑧=πœƒ). Using π‘Ÿ and πœƒ, we can express 𝑧 as 𝑧=4√2ο€»ο€»πœ‹4+π‘–ο€»πœ‹4.cossin

A complex number expressed in this form is said to be in polar form.

Definition: Polar Form of a Complex Number

A complex number 𝑧 written in the form 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ),cossin where the modulus |𝑧|=π‘Ÿ and argument arg𝑧=πœƒ are said to be in polar form. This form is also referred to as the trigonometric form or the modulus–argument form.

Example 1: Recognizing the Polar Form of a Complex Number

Which of the following complex numbers are correctly expressed in polar form?

  1. √2ο€»ο€»πœ‹2+π‘–ο€»πœ‹2sincos
  2. 5ο€Όο€Όβˆ’5πœ‹6+π‘–ο€Όβˆ’5πœ‹6cossin
  3. 𝑒11πœ‹2οˆβˆ’π‘–ο€Ό11πœ‹2cossin
  4. 3πœ‹4ο€»ο€»βˆš35+π‘–ο€»βˆš35cossin
  5. 2ο€Όο€Ό35πœ‹7+𝑖35πœ‹6cossin

Answer

A complex number 𝑧 is said to be in polar form if 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ),cossin where π‘Ÿ is the modulus of 𝑧 and πœƒ is the argument. We will consider each option in turn and see whether it is correctly expressed in this form.

  1. Initially, this number looks like it is in polar form. However, closer inspection reveals that it is not actually in polar form. In fact, it demonstrates one common error students make when writing complex numbers in polar form: getting sine and cosine the wrong way around. To correctly write this number in polar form, we can use the cofunction identities: cossinsincosπœƒ=ο€»πœ‹2βˆ’πœƒο‡,πœƒ=ο€»πœ‹2βˆ’πœƒο‡. Hence, setting πœƒ=0, we can correctly express this number in polar form as √2((0)+𝑖(0)).sincos
  2. This number is in the form π‘Ÿ(πœƒ+π‘–πœƒ)cossin, where π‘Ÿ=5 and πœƒ=βˆ’5πœ‹6. Since π‘Ÿβ‰₯0, this number is correctly written in polar form.
  3. One again, this number is suspiciously close to being in the correct form. However, the minus sign before π‘–πœƒsin means that the number is not in the correct form. To correct this, we can use the even/odd identities, sinsincoscos(βˆ’πœƒ)=βˆ’(πœƒ),(βˆ’πœƒ)=(πœƒ), to rewrite the number as 𝑒11πœ‹2οˆβˆ’π‘–ο€Ό11πœ‹2=𝑒11πœ‹2+π‘–ο€Όβˆ’11πœ‹2=π‘’ο€Όο€Όβˆ’11πœ‹2+π‘–ο€Όβˆ’11πœ‹2,cossincossincossin which is now correctly expressed in polar form.
  4. It would be easy to mistakenly assume that this number it not in polar form since it looks like π‘Ÿ and πœƒ are the wrong way around. However, π‘Ÿ can be any positive real number; hence, 3πœ‹4 a perfectly legitimate value for π‘Ÿ. Similarly, πœƒ can take any real value, which means that √35 is a legitimate value of the argument.
    Hence, this number is correctly written in polar form.
  5. It can be easy to miss the fact that this number is not in polar form. However, a careful look reveals that the sine and cosine functions have different arguments. To rewrite this in the correct form, we would need to calculate its actual modulus and argument.

Therefore, the only two numbers that are correctly expressed in polar form are 5ο€Όο€Όβˆ’5πœ‹6+π‘–ο€Όβˆ’5πœ‹6cossin and 3πœ‹4ο€»ο€»βˆš35+π‘–ο€»βˆš35cossin.

Gaining fluency in converting between the algebraic form (π‘₯+𝑖𝑦) and the polar form of complex numbers will prove to be extremely useful. We will now look at an example of converting a complex number from algebraic to polar form.

Example 2: Converting a Complex Number from Algebraic Form to Polar Form

  1. Find the modulus of the complex number 1+𝑖.
  2. Find the argument of the complex number 1+𝑖.
  3. Hence, write the complex number 1+𝑖 in polar form.

Answer

Part 1

Recall that the modulus of a complex number 𝑧=π‘Ž+𝑖𝑏 is given by |𝑧|=βˆšπ‘Ž+𝑏.

Hence, the modulus of 1+𝑖 is √1+1=√2.

Part 2

When calculating the argument of a complex number, we need to be careful to check in which quadrant of the Argand diagram the complex number lies.

Since we are in the first quadrant, we can just use the inverse tangent function to find the argument. Hence, argarctanarctan(1+𝑖)=ο€Ό11=(1)=πœ‹4.

Part 3

Finally, using the definition of polar form, we can write 1+𝑖=√2ο€»πœ‹4+π‘–πœ‹4.cossin

How to Convert a Complex Number from Algebraic Form to Polar Form

To convert a complex number in algebraic form 𝑧=π‘Ž+𝑏𝑖 to polar form, follow the following steps.

  1. Find the modulus, |𝑧|, of the complex number using the formula |𝑧|=βˆšπ‘Ž+π‘οŠ¨οŠ¨.
  2. Find the argument, arg𝑧, of the complex number. There is more than one technique for finding the argument of a complex number; one such technique is presented here. If 𝑧 is in the first or fourth quadrant of the Argand diagram (π‘Ž>0), we can simply use the inverse tangent function and calculate argarctan𝑧=ο€½π‘π‘Žο‰. However, if the complex number is in the second quadrant (π‘Ž<0 and 𝑏>0), we need to add πœ‹ to the value we get using the inverse tangent function. Hence, argarctan𝑧=ο€½π‘π‘Žο‰+πœ‹. But if the complex number is in the third quadrant (π‘Ž<0 and 𝑏<0), we need to subtract πœ‹ from the value we get using the inverse tangent function. Hence,argarctan𝑧=ο€½π‘π‘Žο‰βˆ’πœ‹. Finally, if the complex number is purely imaginary (π‘Ž=0), then arg𝑧=πœ‹2 if 𝑏>0, and arg𝑧=βˆ’πœ‹2 if 𝑏<0. Note that when π‘Ž=𝑏=0, the argument is undefined.
  3. Write the number in polar form: 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ),cossin

where π‘Ÿ=|𝑧|and πœƒ=𝑧arg.

Now we will look at an example where we cannot use the inverse tangent function to find the argument.

Example 3: Converting a Complex Number from Algebraic Form to Trigonometric Form

Express the complex number 𝑍=4𝑖 in trigonometric form.

Answer

Remember that trigonometric form and polar form are two different names for the same thing. We begin by finding the modulus of the complex number 𝑍. Applying the formula to find the modulus, noting that there is no real part, gives |𝑍|=√0+4=4.

Now we would like to find the argument. Notice that 𝑍 is a purely imaginary number with no real part.

In this case, we are not able to apply the inverse tangent function to calculate the modulus sincewe cannot divide by zero. However, given that all complex numbers that lie on the positive imaginary axis have an argument of πœ‹2, we can conclude that arg𝑍=πœ‹2.

Finally, we can write the number in its trigonometric (or polar) form as follows: 𝑍=4ο€»ο€»πœ‹2+π‘–ο€»πœ‹2.cossin

We will nowconsider an example that highlights the relationship between the polar and algebraic forms of a complex number.

Example 4: Relation between Polar and Algebraic Forms of a Complex Number

Consider the diagram.

  1. Which of the following correctly describes the relationship between π‘Ž, π‘Ÿ and πœƒ?
    1. π‘Ž=π‘Ÿπœƒsin
    2. π‘Ž=πœƒπ‘Ÿcos
    3. π‘Ž=π‘Ÿπœƒtan
    4. π‘Ž=π‘Ÿπœƒcos
    5. π‘Ž=πœƒπ‘Ÿsin
  2. Which of the following correctly describes the relationship between 𝑏, π‘Ÿ and πœƒ?
    1. 𝑏=πœƒπ‘Ÿsin
    2. 𝑏=π‘Ÿπœƒcos
    3. 𝑏=π‘Ÿπœƒsin
    4. 𝑏=π‘Ÿπœƒtan
    5. 𝑏=πœƒπ‘Ÿcos
  3. Hence, express 𝑧 in terms of π‘Ÿ and πœƒ.

Answer

Part 1

The triangle with sides π‘Ÿ, π‘Ž and 𝑏 is a right triangle with hypotenuse π‘Ÿ. Hence, to write a relationship between π‘Ž, π‘Ÿ and πœƒ, we can apply basic trigonometry. The side of length π‘Ž is adjacent to angle πœƒ. Hence, we can use the cosine as follows: cosadjacenthypotenuseπœƒ==π‘Žπ‘Ÿ.

Rearranging, we can write π‘Ž=π‘Ÿπœƒ.cos

Hence, the correct relationship between π‘Ž, π‘Ÿ and πœƒ is (𝑑).

Part 2

Similarly, the side of length 𝑏 is opposite angle πœƒ. Hence, using the sine function, we can write sinoppositehypotenuseπœƒ==π‘π‘Ÿ.

Rearranging, we can write 𝑏=π‘Ÿπœƒ.sin

Hence, the correct relationship between 𝑏, π‘Ÿ, and πœƒ is (𝑐).

Part 3

Since 𝑧=π‘Ž+𝑏𝑖, we can substitute the values of π‘Ž and 𝑏 into this equation to get 𝑧=π‘Ÿπœƒ+π‘Ÿπ‘–πœƒ.cossin

If we factor out the π‘Ÿ in the final part of the previous example, we find that we have an expression for 𝑧 in polar form, whereas in its current form, we have

which is 𝑧 expressed in algebraic form: π‘Ž+𝑏𝑖.

We will now have a look at a further example of converting from polar form to algebraic from.

Example 5: Converting Complex Numbers from Trigonometric to Rectangular Form

  1. Find cosπœ‹6.
  2. Find sinπœ‹6.
  3. Hence, express the complex number 10ο€»πœ‹6+π‘–πœ‹6cossin in rectangular form.

Answer

Part 1

Recall that πœ‹6 is a β€œspecial” angle. Hence, we should have committed its sine, cosine, and tangent values to memory and simply be able to recall cosπœ‹6=√32.

Part 2

Similarly, sinπœ‹6=12.

Part 3

Recall that rectangular form is another name for algebraic form. Hence, using the two values we have just given, we can rewrite the complex number as follows: 10ο€»πœ‹6+π‘–πœ‹6=10ο€Ώβˆš32+12𝑖.cossin

Expanding the brackets, we have 10ο€»πœ‹6+π‘–πœ‹6=5√3+5𝑖.cossin

As we saw in the previous example, converting from polar form to algebraic form is as simple as evaluating sine and cosine and then expanding the brackets.

Example 6: Argument of a Complex Number

Given that 𝑍=πœƒβˆ’π‘–πœƒsincos, where πœƒβˆˆο“0,πœ‹2, find the principal argument of 𝑍.

Answer

Since 𝑍 is given in a form that is similar to polar form, the simplest way to solve this problem will be using trigonometric identities to rewrite 𝑍 in polar form. Once we have 𝑍 in polar form, we can simply read off its argument. First, sine and cosine are the wrong way around for polar form. To resolve this, we can use the cofunction identities, which relate sine and cosine as follows: cossin=ο€»πœ‹2βˆ’πœƒο‡,sincos=ο€»πœ‹2βˆ’πœƒο‡.

Using these, we can rewrite 𝑍 as follows: 𝑍=πœƒβˆ’π‘–πœƒ=ο€»πœ‹2βˆ’πœƒο‡βˆ’π‘–ο€»πœ‹2βˆ’πœƒο‡.sincoscossin

This is not quite in polar form yet: we have a negative sign before sine, whereas, for polar form, it should be positive. To correct this, we can use the even/odd identities, sinsincoscos(βˆ’πœƒ)=βˆ’(πœƒ),(βˆ’πœƒ)=(πœƒ), to rewrite 𝑍 as 𝑍=ο€»πœ‹2βˆ’πœƒο‡+π‘–ο€»βˆ’ο€»πœ‹2βˆ’πœƒο‡ο‡=ο€»βˆ’ο€»πœ‹2βˆ’πœƒο‡ο‡+π‘–ο€»βˆ’ο€»πœ‹2βˆ’πœƒο‡ο‡.cossincossin

Simplifying, we have 𝑍=ο€»πœƒβˆ’πœ‹2+π‘–ο€»πœƒβˆ’πœ‹2.cossin

Now that 𝑍 is expressed in polar form, we can simply read off its argument as πœƒβˆ’πœ‹2. Finally, we need to check whether this is the principal argument. Recall that, for any given argument to be considered the principal argument, it must be given in the range (βˆ’πœ‹,πœ‹]. We are given that πœƒβˆˆο“0,πœ‹2. Hence, 0β‰€πœƒ<πœ‹2.

Subtracting πœ‹2, we get βˆ’πœ‹2β‰€πœƒβˆ’πœ‹2<0.

Hence, arg(𝑍)=πœƒβˆ’πœ‹2 is in the range ο“βˆ’πœ‹2,0, which is contained in (βˆ’πœ‹,πœ‹] and is, therefore, the principal argument of 𝑍.

Key Points

  1. Similar to howpoints in the plane can be written in Cartesian and polar coordinates, we can write complex numbers in algebraic and polar forms.
  2. The polar form of a complex number 𝑧 is 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ),cossin where π‘Ÿβ‰₯0 is the modulus and πœƒ is the argument.
  3. To convert a complex number 𝑧=π‘Ž+𝑏𝑖 to polar form, we simply calculate its modulus π‘Ÿ and its argument πœƒ=(𝑧)arg.
  4. To convert a complex number from polar form 𝑧=π‘Ÿ(πœƒ+π‘–πœƒ)cossin to algebraic form, we can expand the brackets and evaluate sine and cosine.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.