Real and imaginary numbers combine to form complex numbers. The imaginary component, I (iota), indicates a square root of -1. The imaginary portion of a complex number is i. a + ib is a typical representation of complex numbers in their rectangular or standard form. For example, 100 + 25i is a complex number in which 100 represents the real part and 25i represents the imaginary part.
Polar Form Representation of Complex Numbers
To represent a complex number, the polar coordinates of the real and imaginary components are written here. represents the angle at which the number line is inclined to the real axis, i.e. the x-axis. The length indicated by the line is known as its modulus, and it is represented by the letter r in the alphabet. The real and imaginary components are represented by a and b, respectively, while the modulus is represented by OP = r in the diagram below.

The Pythagoras theorem is to be used to get the length r. Trigonometric ratios can be used to calculate arguments. The polar form of a complex number of the type z = a + ib is represented as follows:
r = Modulus[cos(argument) + isin(argument)]
Alternatively, z = r[cosθ + isinθ]
In this case, r =
and θ = tan-1{b/a}.
Calculating Roots of Complex Numbers
DeMoivre’s Theorem can be used to simplify higher-order complex numbers. It can be used to determine the roots of complex numbers as well as expand complex numbers according to their exponent.
Given:
, then its roots are:
![Rendered by QuickLaTeX.com r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-dfb5fde62c6b35f34ffce445ce04eadb_l3.png)
Where, k lies between 0 and n – 1 and n is the exponent or radical.
Find the fourth roots of 5(1 + i√3). Leave in trigonometric form.
Solution:
Modulus of the given number =
= 10
Argument = tan-1[5√3/ 5] = π/3.
Thus, the polar form of 5 + 5√3i = ![Rendered by QuickLaTeX.com 10 [\cos{(\frac{\pi}{3})} + i \sin{(\frac{\pi}{3}})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-7d8cc2e5617082748989a0771f0363ce_l3.png)
According to DeMoivre’s formula, all the nth roots of a complex number are given by:
, where k lies between 0 and n – 1 and n is the exponent or radical.
Here, r = 10, θ = π/3 and n = 4.
Find the 4 roots by substituting the values of k as 0, 1, 2 and 3 respectively.
- For k = 0, z =
![Rendered by QuickLaTeX.com \sqrt[4]{10}[cos(\frac{\frac{\pi}{3}+2\pi(0)}{4})+i\sin(\frac{\frac{\pi}{3}+2\pi(0)}{4})](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-22cee4841ff5f08411197df6869c0bb2_l3.png)
= ![Rendered by QuickLaTeX.com \sqrt[4]{10}[cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-a6f5abcfc1156ecedbe3dd25c78d5965_l3.png)
- For k = 1, z =
![Rendered by QuickLaTeX.com \sqrt[4]{10}[cos(\frac{\frac{\pi}{3}+2\pi(1)}{4})+i\sin(\frac{\frac{\pi}{3}+2\pi(1)}{4})](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-346516f567d6f9837ab7d294b4f90ecb_l3.png)
= ![Rendered by QuickLaTeX.com \sqrt[4]{10}[cos(\frac{7\pi}{12})+i\sin(\frac{7\pi}{12})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-ce1c88ba6ae539d65f4a5cb7d40d3b7f_l3.png)
- For k = 2, z =
![Rendered by QuickLaTeX.com \sqrt[4]{10} (\cos{(\frac{\frac{\pi}{3} + 2\cdot \pi\cdot 2}{4} )} + i \sin{(\frac{\frac{\pi}{3} + 2\cdot \pi\cdot 2}{4})})](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-0e354ac58b4609cfdedac669733483e9_l3.png)
= ![Rendered by QuickLaTeX.com \sqrt[4]{10}[cos(\frac{13\pi}{12})+i\sin(\frac{13\pi}{12})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-311c7ba12713c0af8c90768f8bdc78c9_l3.png)
- For k = 3, z =
![Rendered by QuickLaTeX.com \sqrt[4]{10}[cos(\frac{\frac{\pi}{3}+2\pi(3)}{4})+i\sin(\frac{\frac{\pi}{3}+2\pi(3)}{4})](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-db35c3b5e6ca3c0bea8acc0a9b4bd678_l3.png)
= ![Rendered by QuickLaTeX.com \sqrt[4]{10}[cos(\frac{19\pi}{12})+i\sin(\frac{19\pi}{12})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-27f987dc0b1d7dd0b64408e8670da50b_l3.png)
Thus, the four roots of 5(1 + i√3) are
and
.
Similar Problems
Question 1: Find the cube roots of -2 – 2√3i.
Solution:
r =
= √(16) = 4, θ = 4π/ 3.
According to DeMoivre’s formula, all the nth roots of a complex number are given by:
, where k lies between 0 and n – 1 and n is the exponent or radical.
Find the 3 roots by substituting the values of m as 0, 1 and 2 respectively,
For m = 0, z = ![Rendered by QuickLaTeX.com \sqrt[3]4.[cos(\frac{\frac{4\pi}{3}+2\pi(0)}{3})+i\ sin(\frac{\frac{4\pi}{3}+2π(0)}{3})]\\ =\sqrt[3]4.[cos(\frac{4\pi}{9})+i\ sin(\frac{4\pi}{9})]\\ =\sqrt[3]4.[0.173+0.98i]\\](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-bd2613014a5607cb234cdd78f602a810_l3.png)
= 0.27 + 1.56i
For m = 1, z = ![Rendered by QuickLaTeX.com \sqrt[3]4.[cos(\frac{\frac{4\pi}{3}+2\pi(1)}{3})+i\ sin(\frac{\frac{4\pi}{3}+2\pi(1)}{3})]\\ =\sqrt[3]4.[cos(\frac{10\pi}{9})+i\ sin(\frac{10\pi}{9})]\\ =\sqrt[3]4.[cos(\pi+\frac{\pi}{9})+i\ sin(\pi+\frac{\pi}{9})]\\ =\sqrt[3]4.[-cos(\frac{\pi}{9})-i\ sin(\frac{\pi}{9})]\\ =\sqrt[3]4.[-0.93-0.34i]\\](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-d47960124817f8f43932b5d7d1abc4f7_l3.png)
= −1.49 − 0.54i
For m = 2, z = ![Rendered by QuickLaTeX.com \sqrt[3]4.[cos(\frac{\frac{4\pi}{3}+2\pi(2)}{3})+i\ sin(\frac{\frac{4\pi}{3}+2\pi(2)}{3})]\\ =\sqrt[3]4.[cos(\frac{16\pi}{9})+i\ sin(\frac{16\pi}{9})]\\ =\sqrt[3]4.[0.76-0.64i]\\](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-00636cf7f5784aa0550f46cbab7970fb_l3.png)
= 1.21 −1.02i
Thus, the roots are 0.27 + 1.56i, −1.49 − 0.54i and 1.21 – 1.02i.
Question 2: Find the fifth roots of 32 + 0i.
Solution:
Modulus =
= 32.
Argument = θ = tan-1(0/ 32) = 0.
According to DeMoivre’s formula, all the nth roots of a complex number are given by:
, where k lies between 0 and n – 1 and n is the exponent or radical.
Find the 5 roots by substituting the values of m as 0, 1, 2, 3 and 4.
For m = 0,
![Rendered by QuickLaTeX.com z = 2[cos(\frac{2π(0)}{5})+i\ sin(\frac{2π(0)}{5})\\ =2[cos(0)+i\ sin(0)]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-b4f6d139bc4ad248d5dad540440ca782_l3.png)
= 2
For m = 1,
z =
= 0.62 + 1.9i
For m = 2,
z =
= −1.62 + 1.18i
For m = 3,
![Rendered by QuickLaTeX.com 2[cos(\frac{2π(3)}{5})+i\ sin(\frac{2π(3)}{5})\\ =2[cos(\frac{6π)}{5})+i\ sin(\frac{6π}{5})]\\ = 2 [-0.80-0.58i]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-d6a723f171e7176a4dd9350179f4b821_l3.png)
= −1.62 − 1.18i
For m = 4,
![Rendered by QuickLaTeX.com [cos(\frac{2π(4)}{5})+i\ sin(\frac{2π(4)}{5})\\ =2[cos(\frac{8π)}{5})+i\ sin(\frac{8π}{5})]\\ = 2 [-0.30-0.95i]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-5814b43878e2f270f0e74fe1f19f65f8_l3.png)
= 0.62 − 1.9i
Thus, the roots are 2, 0.62 + 1.9i, -1.62 + 1.18i, -1.62 – 1.18i and 0.62 – 1.9i.
Question 3: Find the fourth roots of -8√3 + 8i.
Solution:
Polar form = ![Rendered by QuickLaTeX.com 16[cos(\frac{5\pi}{6})+isin(\frac{5\pi}{6})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-71e45988edea32bbb28ce9c9f3bcff46_l3.png)
We have k = 2, n = 4 and θ = 5π/ 6.
According to DeMoivre’s formula, all the nth roots of a complex number are given by:
, where k lies between 0 and n – 1 and n is the exponent or radical.
Find the 4 roots by substituting the values of m as 0, 1, 2 and 3 respectively.
For m = 0,
![Rendered by QuickLaTeX.com z = 2[cos(\frac{\frac{5π}{6}+2π(0)}{4})+i\ sin(\frac{\frac{5π}{6}+2π(0)}{4})\\ =2[cos(\frac{5\pi}{24})+i\ sin(\frac{5\pi}{24})]\\ =2[0.79+0.60i]\\ = 1.58 + 1.21i](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-691112f19983ecb86ca707ca82da1294_l3.png)
For m = 1,
![Rendered by QuickLaTeX.com z = 2[cos(\frac{\frac{5π}{6}+2π(1)}{4})+i\ sin(\frac{\frac{5π}{6}+2π(1)}{4})\\ =2[cos(\frac{17π}{24})+i\ sin(\frac{17π}{24})\\ =2[-0.60+0.79i]\\ = −1.21 + 1.58i](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-531115917d6b71baef689e6f572c15ed_l3.png)
For m = 2,
![Rendered by QuickLaTeX.com z = 2[cos(\frac{\frac{5π}{6}+2π(2)}{4})+i\ sin(\frac{\frac{5π}{6}+2π(2)}{4})\\ =2[cos(\frac{29π}{24})+i\ sin(\frac{29π}{24})\\ =2[-0.79-0.60i]\\ =−1.58 −1.21i](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-d536ad0661b28f8910924e2c95f78efe_l3.png)
For m = 3,
![Rendered by QuickLaTeX.com z = 2[cos(\frac{\frac{5π}{6}+2π(3)}{4})+i\ sin(\frac{\frac{5π}{6}+2π(3)}{4})\\ =2[cos(\frac{41π}{24})+i\ sin(\frac{41π}{24})\\ =2[0.60-0.79i]\\ =1.21 − 1.58i](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-496e3b44c4b6187ae3f90a65450f6127_l3.png)
Thus, the four roots of z are 1.58 + 1.21i, −1.21 + 1.58i, −1.58 −1.21i and 1.21 − 1.58i.
Question 4: Find the sixth root of -27i. Leave in trigonometric form.
Solution:
Modulus = 
Argument = θ = tan-1(-27/ 0) = π/2.
Polar form = ![Rendered by QuickLaTeX.com 27[cos(\frac{-π}{2})+isin(\frac{-π}{2})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-6fbe854cc5d879b9dcec3094d27ea0f0_l3.png)
According to DeMoivre’s formula, all the nth roots of a complex number are given by:
, where k lies between 0 and n – 1 and n is the exponent or radical.
Find the 6 roots by substituting the values of m as 0, 1, 2, 3, 4 and 5.
- For m = 0, z =
![Rendered by QuickLaTeX.com \sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2π(\theta)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(\theta)}{6})](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-8d62a01f7b57c6f3f2e921a770bb147e_l3.png)
= ![Rendered by QuickLaTeX.com \sqrt{3}[cos(\frac{-\pi}{12})+i\ sin(\frac{-\pi}{12})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-bc7c10b60f761ab052b75c2734e20cd3_l3.png)
- For m = 1, z =
![Rendered by QuickLaTeX.com \sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(1)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(1)}{6})](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-1f6d471585e2191f21f7fe6ae5c3393d_l3.png)
= ![Rendered by QuickLaTeX.com \sqrt{3}[cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-dd5feffa3f4bbb1528e8422ec2d213e5_l3.png)
- For m = 2, z =
![Rendered by QuickLaTeX.com \sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(2)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(2)}{6})](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-2ed22e7b0130dc40cbad2ed116dca9bb_l3.png)
= ![Rendered by QuickLaTeX.com \sqrt{3}[cos(\frac{7\pi}{12})+i\ sin(\frac{7\pi}{12})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-0377cd0ed671f7c7e6371f0b39d45515_l3.png)
- For m = 3, z =
![Rendered by QuickLaTeX.com \sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(3)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(3)}{6})](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-f233d3c49045f203190fc9f80fc74ed3_l3.png)
= ![Rendered by QuickLaTeX.com \sqrt{3}[cos(\frac{11\pi}{12})+i\ sin(\frac{11\pi}{12})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-a2c550d0c04d044d71f380d6481b43e8_l3.png)
- For m = 4, z =
![Rendered by QuickLaTeX.com \sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(4)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(4)}{6})](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-c11940cb2b978d7dfca21fc0537407a9_l3.png)
= ![Rendered by QuickLaTeX.com \sqrt{3}[cos(\frac{5\pi}{4})+i\ sin(\frac{5\pi}{4})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-74b5c4c389375b416af8be1f751aff7f_l3.png)
- For m = 5, z =
![Rendered by QuickLaTeX.com \sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(5)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(5)}{6})](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-e7a92eadeebe53c5efa6417f8b68e6bd_l3.png)
= ![Rendered by QuickLaTeX.com \sqrt{3}[cos(\frac{19\pi}{12})+i\ sin(\frac{19\pi}{12})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-37af7701b4968c27cd25283e129f4f85_l3.png)
Question 5: Find the fourth root of 81i. Leave in trigonometric form.
Solution:
Polar form = ![Rendered by QuickLaTeX.com 81[cos(\frac{\pi}{2})+isin(\frac{\pi}{2})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-a70e54d2d785e473f052d6ae028906ec_l3.png)
We have k = 81, n = 4 and θ = π/ 2.
According to DeMoivre’s formula, all the nth roots of a complex number are given by:
, where k lies between 0 and n – 1 and n is the exponent or radical.
Find the 4 roots by substituting the values of m as 0, 1, 2 and 3 respectively.
- For m = 0, z =
= ![Rendered by QuickLaTeX.com 3[cos(\frac{\pi}{8})+i\ sin(\frac{\pi}{8})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-e1f2a70a3bf883a87b6f8ca4baa767df_l3.png)
- For m = 1, z =
= ![Rendered by QuickLaTeX.com 3[cos(\frac{5\pi}{8})+i\ sin(\frac{5\pi}{8})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-4e66f0af407b16a8374550ebdaaee6cb_l3.png)
- For m = 2, z =
= ![Rendered by QuickLaTeX.com 3[cos(\frac{9\pi}{8})+i\ sin(\frac{9\pi}{8})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-e4e474d217cd7631183ddb0cd65804fe_l3.png)
- For m = 3, z =
![Rendered by QuickLaTeX.com \sqrt[4]{81}[cos(\frac{\frac{\pi}{2}+2\pi(3)}{4})+i\ sin(\frac{\frac{\pi}{2}+2\pi(3)}{4}) = 3[cos(\frac{13\pi}{8})+i\ sin(\frac{13\pi}{8})]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-efe2fe67229cbeb361b49b9c72d39727_l3.png)