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Find all the complex cube roots of w = 8 (cos 150° + i sin 150°)

  • Last Updated : 18 Jan, 2022

Complex numbers are the numbers of the form a + ib, such that a and b are real numbers and i (iota) is the imaginary component and represents √(-1), commonly depicted in their rectangular or standard form. For example, 10 + 5i is a complex number where 10 is the real part and 5i is the imaginary part. These can be purely real or purely imaginary depending upon the values of a and b. If a = 0 in a + ib, then ib is a purely imaginary number, and if b = 0, then we have a, which is a purely real number.

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 + i\ sinθ) , 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 all the complex cube roots of w = 8 (cos 150° + i sin 150°)

Solution:

w = 8(Cos 150° + i sin 150°)

The above complex number can also be expressed as w = 8(cos(150° + 360n) + i sin(150° + 360n).

\sqrt[3]{w} = \sqrt[3]{8[(cos(150° + 360n) + i sin(150° + 360n)]}

w1/3 = {8[(cos(150° + 360n) + i sin(150° + 360n)]}1/3

= 2(3)(1/3) [(cos(150° + 360n) + i sin(150° + 360n)]1/3

As per DeMoivre’s Theorem, (cos x + isinx)n = cos(nx) + isin(nx). 

2[(cos(\frac{150°+360n}{3}) + i sin(\frac{150°+360n}{3})]^{1/3}

Substitute n = 0, 1, 2 to find the roots.

  • For n = 0, w12[(cos(\frac{150°+360(0)}{3}) + i sin(\frac{150°+360(0)}{3})]^{1/3}   = 2(cos 50° + i sin 50°)
  • For n = 1, w22[(cos(\frac{150°+360(1)}{3}) + i sin(\frac{150°+360(1)}{3})]^{1/3}   = 2(cos 170° + i sin 170°)
  • For n = 2, w32[(cos(\frac{150°+360(2)}{3}) + i sin(\frac{150°+360(2)}{3})]^{1/3}   = 2(cos 290° + i sin 290°)

Similar Problems

Question 1. Find all the complex cube roots of w = 125(Cos 150° + i sin 150°). Write the roots in polar form with theta in degrees.

Solution:

w = 125(Cos 150° + i sin 150°)

The above complex number can also be expressed as w = 125 (cos(150° + 360n) + i sin(150° + 360n).

\sqrt[3]{w} = \sqrt[3]{125[(cos(150° + 360n) + i sin(150° + 360n)]}

w1/3{125[(cos(150° + 360n) + i sin(150° + 360n)]}^{1/3}

= 5(3)(1/3) [(cos(150° + 360n) + i sin(150° + 360n)]^{1/3}

As per DeMoivre’s Theorem, (cos x + isinx)n = cos(nx) + isin(nx). 

5[(cos(\frac{150°+360n}{3}) + i sin(\frac{150°+360n}{3})]^{1/3}

Substitute n = 0,1,2 to find the roots.

  • For n = 0, w15[(cos(\frac{150°+360(0)}{3}) + i sin(\frac{150°+360(0)}{3})]^{1/3}   = 2(cos 50° + i sin 50°)
  • For n = 1, w25[(cos(\frac{150°+360(1)}{3}) + i sin(\frac{150°+360(1)}{3})]^{1/3}   = 2(cos 170° + i sin 170°)
  • For n = 2, w35[(cos(\frac{150°+360(2)}{3}) + i sin(\frac{150°+360(2)}{3})]^{1/3}   = 2(cos 290° + i sin 290°)

Question 2. Find all the complex cube roots of w = 27(Cos 150° + i sin 150°). Write the roots in polar form with theta in degrees.

Solution:

w = 27(Cos 150° + i sin 150°)

The above complex number can also be expressed as w = 27(cos(150° + 360n) + i sin(150° + 360n).

\sqrt[3]{w} = \sqrt[3]{27[(cos(150° + 360n) + i sin(150° + 360n)]}

w1/3{27[(cos(150° + 360n) + i sin(150° + 360n)]}^{1/3}

= 3(3)(1/3) [(cos(150° + 360n) + i sin(150° + 360n)]^{1/3}

As per DeMoivre’s Theorem, (cos x + isinx)n = cos(nx) + isin(nx). 

3[(cos(\frac{150°+360n}{3}) + i sin(\frac{150°+360n}{3})]^{1/3}

Substitute n = 0,1,2 to find the roots.

  • For n = 0, w13[(cos(\frac{150°+360(0)}{3}) + i sin(\frac{150°+360(0)}{3})]^{1/3}   = 2(cos 50° + i sin 50°)
  • For n = 1, w23[(cos(\frac{150°+360(1)}{3}) + i sin(\frac{150°+360(1)}{3})]^{1/3}   = 3(cos 170° + i sin 170°)
  • For n = 2, w33[(cos(\frac{150°+360(2)}{3}) + i sin(\frac{150°+360(2)}{3})]^{1/3}   = 3(cos 290° + i sin 290°)

Question 3. Find all the complex cube roots of w = 64 (Cos 150° + i sin 150°). Write the roots in polar form with theta in degrees.

Solution:

w = 64(Cos 150° + i sin 150°)

The above complex number can also be expressed as w = 64(cos(150° + 360n) + i sin(150° + 360n).

\sqrt[3]{w} = \sqrt[3]{64[(cos(150° + 360n) + i sin(150° + 360n)]}

w1/3{64[(cos(150° + 360n) + i sin(150° + 360n)]}1/3

= 4(3)(1/3) [(cos(150° + 360n) + i sin(150° + 360n)]^{1/3}

As per DeMoivre’s Theorem, (cos x + isinx)n = cos(nx) + isin(nx). 

4[(cos(\frac{150°+360n}{3}) + i sin(\frac{150°+360n}{3})]^{1/3}

Substitute n = 0,1,2 to find the roots.

  • For n = 0, w1 = 4[(cos(\frac{150°+360(0)}{3}) + i sin(\frac{150°+360(0)}{3})]^{1/3}   = 4(cos 50° + i sin 50°)
  • For n = 1, w2 = 4[(cos(\frac{150°+360(1)}{3}) + i sin(\frac{150°+360(1)}{3})]^{1/3}   = 4(cos 170° + i sin 170°)
  • For n = 2, w3 = 4[(cos(\frac{150°+360(2)}{3}) + i sin(\frac{150°+360(2)}{3})]^{1/3}   = 4(cos 290° + i sin 290°)

Question 4. Find all the complex cube roots of w = 343 (Cos 150° + i sin 150°). Write the roots in polar form with theta in degrees.

Solution:

w = 343(Cos 150° + i sin 150°)

The above complex number can also be expressed as w = 343(cos(150° + 360n) + i sin(150° + 360n).

\sqrt[3]{w} = \sqrt[3]{343[(cos(150° + 360n) + i sin(150° + 360n)]}

w1/3{343[(cos(150° + 360n) + i sin(150° + 360n)]}^{1/3}

= 7(3)(1/3) [(cos(150° + 360n) + i sin(150° + 360n)]^{1/3}

As per DeMoivre’s Theorem, (cos x + isinx)n = cos(nx) + isin(nx). 

7[(cos(\frac{150°+360n}{3}) + i sin(\frac{150°+360n}{3})]^{1/3}

Substitute n = 0,1,2 to find the roots.

  • For n = 0, w1 = 7[(cos(\frac{150°+360(0)}{3}) + i sin(\frac{150°+360(0)}{3})]^{1/3}   = 7(cos 50° + i sin 50°)
  • For n = 1, w2 = 7[(cos(\frac{150°+360(1)}{3}) + i sin(\frac{150°+360(1)}{3})]^{1/3}   = 7(cos 170° + i sin 170°)
  • For n = 2, w3 = 7[(cos(\frac{150°+360(2)}{3}) + i sin(\frac{150°+360(2)}{3})]^{1/3}   = 7(cos 290° + i sin 290°)

Question 5. Find all the complex cube roots of w = 729 (Cos 150° + i sin 150°). Write the roots in polar form with theta in degrees.

Solution:

w = 729(Cos 150° + i sin 150°)

The above complex number can also be expressed as w = 729(cos(150° + 360n) + i sin(150° + 360n).

\sqrt[3]{w} = \sqrt[3]{729[(cos(150° + 360n) + i sin(150° + 360n)]}

w1/3{729[(cos(150° + 360n) + i sin(150° + 360n)]}^{1/3}

= 9(3)(1/3) [(cos(150° + 360n) + i sin(150° + 360n)]^{1/3}

As per DeMoivre’s Theorem, (cos x + isinx)n = cos(nx) + isin(nx). 

9[(cos(\frac{150°+360n}{3}) + i sin(\frac{150°+360n}{3})]^{1/3}

Substitute n = 0,1,2 to find the roots.

  • For n = 0, w1 = 9[(cos(\frac{150°+360(0)}{3}) + i sin(\frac{150°+360(0)}{3})]^{1/3}   = 9(cos 50° + i sin 50°)
  • For n = 1, w2 = 9[(cos(\frac{150°+360(1)}{3}) + i sin(\frac{150°+360(1)}{3})]^{1/3 }  = 9(cos 170° + i sin 170°)
  • For n = 2, w3 = 9[(cos(\frac{150°+360(2)}{3}) + i sin(\frac{150°+360(2)}{3})]^{1/3}   = 9(cos 290° + i sin 290°)

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