Find the fourth roots of 5(1 + i√3)

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 = $\sqrt{a^2+b^2}$ 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: $z^{\frac{1}{n}} = r^{\frac{1}{n}}(cosθ + isinθ)$ , then its roots are:

$r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]$

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 = $\sqrt{5^2+(5\sqrt{3})^2}$ = 10
Argument = tan^-1[5√3/ 5] = π/3.
Thus, the polar form of 5 + 5√3i = $10 [\cos{(\frac{\pi}{3})} + i \sin{(\frac{\pi}{3}})]$
According to DeMoivre’s formula, all the nth roots of a complex number are given by:
$r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]$ , 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 = $\sqrt[4]{10}[cos(\frac{\frac{\pi}{3}+2\pi(0)}{4})+i\sin(\frac{\frac{\pi}{3}+2\pi(0)}{4})$
= $\sqrt[4]{10}[cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12})]$
For k = 1, z = $\sqrt[4]{10}[cos(\frac{\frac{\pi}{3}+2\pi(1)}{4})+i\sin(\frac{\frac{\pi}{3}+2\pi(1)}{4})$
= $\sqrt[4]{10}[cos(\frac{7\pi}{12})+i\sin(\frac{7\pi}{12})]$
For k = 2, z = $\sqrt[4]{10} (\cos{(\frac{\frac{\pi}{3} + 2\cdot \pi\cdot 2}{4} )} + i \sin{(\frac{\frac{\pi}{3} + 2\cdot \pi\cdot 2}{4})})$
= $\sqrt[4]{10}[cos(\frac{13\pi}{12})+i\sin(\frac{13\pi}{12})]$
For k = 3, z = $\sqrt[4]{10}[cos(\frac{\frac{\pi}{3}+2\pi(3)}{4})+i\sin(\frac{\frac{\pi}{3}+2\pi(3)}{4})$
= $\sqrt[4]{10}[cos(\frac{19\pi}{12})+i\sin(\frac{19\pi}{12})]$
Thus, the four roots of 5(1 + i√3) are $\sqrt[4]{10}[cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12})], \sqrt[4]{10}[cos(\frac{7\pi}{12})+i\sin(\frac{7\pi}{12})], \sqrt[4]{10}[cos(\frac{13\pi}{12})+i\sin(\frac{13\pi}{12})]$ and $\sqrt[4]{10}[cos(\frac{19\pi}{12})+i\sin(\frac{19\pi}{12})]$ .

Similar Problems

Question 1: Find the cube roots of -2 – 2√3i.

Solution:

r = $\sqrt{x^2+y^2}$ = √(16) = 4, θ = 4π/ 3.
According to DeMoivre’s formula, all the nth roots of a complex number are given by:
$r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]$ , 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 = $\sqrt[3]4.[cos(\frac{\frac{4\pi}{3}+2\pi(0)}{3})+i\ sin(\frac{\frac{4\pi}{3}+2π(0)}{3})]\\ =\sqrt[3]4.[cos(\frac{4\pi}{9})+i\ sin(\frac{4\pi}{9})]\\ =\sqrt[3]4.[0.173+0.98i]\\$
= 0.27 + 1.56i
For m = 1, z = $\sqrt[3]4.[cos(\frac{\frac{4\pi}{3}+2\pi(1)}{3})+i\ sin(\frac{\frac{4\pi}{3}+2\pi(1)}{3})]\\ =\sqrt[3]4.[cos(\frac{10\pi}{9})+i\ sin(\frac{10\pi}{9})]\\ =\sqrt[3]4.[cos(\pi+\frac{\pi}{9})+i\ sin(\pi+\frac{\pi}{9})]\\ =\sqrt[3]4.[-cos(\frac{\pi}{9})-i\ sin(\frac{\pi}{9})]\\ =\sqrt[3]4.[-0.93-0.34i]\\$
= −1.49 − 0.54i
For m = 2, z = $\sqrt[3]4.[cos(\frac{\frac{4\pi}{3}+2\pi(2)}{3})+i\ sin(\frac{\frac{4\pi}{3}+2\pi(2)}{3})]\\ =\sqrt[3]4.[cos(\frac{16\pi}{9})+i\ sin(\frac{16\pi}{9})]\\ =\sqrt[3]4.[0.76-0.64i]\\$
= 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 = $\sqrt{x^2+y^2} = \sqrt{(32)^2}$ = 32.
Argument = θ = tan^-1(0/ 32) = 0.
According to DeMoivre’s formula, all the nth roots of a complex number are given by:
$r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]$ , 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,
$z = 2[cos(\frac{2π(0)}{5})+i\ sin(\frac{2π(0)}{5})\\ =2[cos(0)+i\ sin(0)]$
= 2
For m = 1,
z = $2[cos(\frac{2π(1)}{5})+i\ sin(\frac{2π(1)}{5})\\ =2[0.30+0.95]$
= 0.62 + 1.9i
For m = 2,
z = $2[cos(\frac{2π(2)}{5})+i\ sin(\frac{2π(2)}{5})\\ =2[cos(\frac{4π)}{5})+i\ sin(\frac{4π}{5})]\\ = 2 [-0.80+0.58i]$
= −1.62 + 1.18i
For m = 3,
$2[cos(\frac{2π(3)}{5})+i\ sin(\frac{2π(3)}{5})\\ =2[cos(\frac{6π)}{5})+i\ sin(\frac{6π}{5})]\\ = 2 [-0.80-0.58i]$
= −1.62 − 1.18i
For m = 4,
$[cos(\frac{2π(4)}{5})+i\ sin(\frac{2π(4)}{5})\\ =2[cos(\frac{8π)}{5})+i\ sin(\frac{8π}{5})]\\ = 2 [-0.30-0.95i]$
= 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 = $16[cos(\frac{5\pi}{6})+isin(\frac{5\pi}{6})]$
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:
$r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]$ , 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,
$z = 2[cos(\frac{\frac{5π}{6}+2π(0)}{4})+i\ sin(\frac{\frac{5π}{6}+2π(0)}{4})\\ =2[cos(\frac{5\pi}{24})+i\ sin(\frac{5\pi}{24})]\\ =2[0.79+0.60i]\\ = 1.58 + 1.21i$
For m = 1,
$z = 2[cos(\frac{\frac{5π}{6}+2π(1)}{4})+i\ sin(\frac{\frac{5π}{6}+2π(1)}{4})\\ =2[cos(\frac{17π}{24})+i\ sin(\frac{17π}{24})\\ =2[-0.60+0.79i]\\ = −1.21 + 1.58i$
For m = 2,
$z = 2[cos(\frac{\frac{5π}{6}+2π(2)}{4})+i\ sin(\frac{\frac{5π}{6}+2π(2)}{4})\\ =2[cos(\frac{29π}{24})+i\ sin(\frac{29π}{24})\\ =2[-0.79-0.60i]\\ =−1.58 −1.21i$
For m = 3,
$z = 2[cos(\frac{\frac{5π}{6}+2π(3)}{4})+i\ sin(\frac{\frac{5π}{6}+2π(3)}{4})\\ =2[cos(\frac{41π}{24})+i\ sin(\frac{41π}{24})\\ =2[0.60-0.79i]\\ =1.21 − 1.58i$
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 = $\sqrt{x^2+y^2} = \sqrt{(-32)^2} = 27.$
Argument = θ = tan^-1(-27/ 0) = π/2.
Polar form = $27[cos(\frac{-π}{2})+isin(\frac{-π}{2})]$
According to DeMoivre’s formula, all the nth roots of a complex number are given by:
$r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]$ , 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 = $\sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2π(\theta)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(\theta)}{6})$
= $\sqrt{3}[cos(\frac{-\pi}{12})+i\ sin(\frac{-\pi}{12})]$
For m = 1, z = $\sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(1)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(1)}{6})$
= $\sqrt{3}[cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]$
For m = 2, z = $\sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(2)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(2)}{6})$
= $\sqrt{3}[cos(\frac{7\pi}{12})+i\ sin(\frac{7\pi}{12})]$
For m = 3, z = $\sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(3)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(3)}{6})$
= $\sqrt{3}[cos(\frac{11\pi}{12})+i\ sin(\frac{11\pi}{12})]$
For m = 4, z = $\sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(4)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(4)}{6})$
= $\sqrt{3}[cos(\frac{5\pi}{4})+i\ sin(\frac{5\pi}{4})]$
For m = 5, z = $\sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(5)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(5)}{6})$
= $\sqrt{3}[cos(\frac{19\pi}{12})+i\ sin(\frac{19\pi}{12})]$

Question 5: Find the fourth root of 81i. Leave in trigonometric form.

Solution:

Polar form = $81[cos(\frac{\pi}{2})+isin(\frac{\pi}{2})]$
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:
$r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]$ , 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 = $\sqrt[4]{81}[cos(\frac{\frac{\pi}{2}+2\pi(\theta)}{4})+i\ sin(\frac{\frac{\pi}{2}+2\pi(\theta)}{4})$ = $3[cos(\frac{\pi}{8})+i\ sin(\frac{\pi}{8})]$
For m = 1, z = $\sqrt[4]{81}[cos(\frac{\frac{\pi}{2}+2\pi(1)}{4})+i\ sin(\frac{\frac{\pi}{2}+2\pi(1)}{4})$ = $3[cos(\frac{5\pi}{8})+i\ sin(\frac{5\pi}{8})]$
For m = 2, z = $\sqrt[4]{81}[cos(\frac{\frac{\pi}{2}+2\pi(2)}{4})+i\ sin(\frac{\frac{\pi}{2}+2\pi(2)}{4})$ = $3[cos(\frac{9\pi}{8})+i\ sin(\frac{9\pi}{8})]$
For m = 3, z = $\sqrt[4]{81}[cos(\frac{\frac{\pi}{2}+2\pi(3)}{4})+i\ sin(\frac{\frac{\pi}{2}+2\pi(3)}{4}) = 3[cos(\frac{13\pi}{8})+i\ sin(\frac{13\pi}{8})]$