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Find the fourth roots of 5(1 + i√3)

  • Last Updated : 31 Dec, 2021

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 trigonometri

rm.

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})]="292" style="vertical-align: -9px;"/>

  • 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(\theta)}{3})+i\ sin(\frac{\frac{4\pi}{3}+2π(\theta)}{3})]

= 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})]

= −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})]

= 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 = [cos(\frac{2\pi(0)}{5})+i\ sin(\frac{2\pi(0)}{5}) = 2
  • For m = 1,  z = [cos(\frac{2\pi(1)}{5})+i\ sin(\frac{2\pi(1)}{5}) = 0.62 + 1.9i
  • For m = 2,  z = [cos(\frac{2\pi(2)}{5})+i\ sin(\frac{2\pi(2)}{5}) = −1.62 + 1.18i
  • For m = 3,  z = [cos(\frac{2\pi(3)}{5})+i\ sin(\frac{2\pi(3)}{5})  = −1.62 − 1.18i
  • For m = 4,  z = [cos(\frac{2\pi(4)}{5})+i\ sin(\frac{2\pi(4)}{5})  = 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{5\pi}{6}+2\pi(\theta)}{4})+i\ sin(\frac{\frac{5\pi}{6}+2\pi(0)}{4})     = 1.58 + 1.21i.
  • For m = 1, z = 2[cos(\frac{\frac{5\pi}{6}+2\pi(1)}{4})+i\ sin(\frac{\frac{5\pi}{6}+2\pi(1)}{4})     = −1.21 + 1.58i.
  • For m = 2, z = 2[cos(\frac{\frac{5\pi}{6}+2\pi(2)}{4})+i\ sin(\frac{\frac{5\pi}{6}+2\pi(2)}{4})     = −1.58 −1.21i.
  • For m = 3, z = 2[cos(\frac{\frac{5\pi}{6}+2\pi(3)}{4})+i\ sin(\frac{\frac{5\pi}{6}+2\pi(3)}{4})     = 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.<

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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})]="289" style="vertical-align: -9px;"/>

  • 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})]

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