Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

How to use DeMoivre’s Theorem to simplify [2(cos π/3 + i sin π/3)]5?

  • Last Updated : 21 Dec, 2021

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.

Polar Form of a Complex Number

Here, the polar coordinates of the real and imaginary parts are written to depict a complex number. The angle at which the number line is inclined to the real axis, i.e., the x-axis, is represented by θ. The length represented by the line is called its modulus and is depicted by the alphabet r. The figure below depicts a and b as the real and imaginary components respectively and OP = r is the modulus.

Clearly, the Pythagoras theorem can be applied for the calculation of the length r. Arguments can be computed using trigonometric ratios. Thus, for a complex number of the form z = p + iq, its polar form is written as follows:

r = Modulus[cos(argument) + isin(argument)]

Or, z = r[cosθ + isinθ]

Here, r = \sqrt{p^2+q^2}  and θ = tan-1{q/p}.

DeMoivre’s Theorem

Basically, a polar form is just another way of representing a given complex number in case its index is 1. In case the exponent of a given complex number exceeds 1, it needs to be evaluated/expanded, this is when DeMoivre’s theorem comes into the picture. In order to expand a complex number as per its given exponent, it first needs to be converted into its polar form, which uses its modulus and argument as its constituents. Then DeMoivre’s theorem is applied, which states the following,

Formula

For all real values of say, a number x,

(cos x + isinx)n = cos(nx) + isin(nx),

Where n can assume any rational value.

How to use DeMoivre’s Theorem to simplify [2(cos π/3 + i sin π/3)]5?

Solution:

As per DeMoivre’s theorem: (cosθ + sinθ)n = cos(nθ) + isin(nθ).

[2(cos(\frac{\pi}{3})+ i\ sin(\frac{\pi}{3})]^5 = 2^5[cos(\frac{5\pi}{3})+i\ sin(\frac{5\pi}{3})]

2^5(\frac{1}{2}-i\frac{\sqrt{3}}{2})

= 24(1 – i√3)

Hence, [2(cos pi/3 + i sin pi/3)]5 = 24 – 24√3i.

Similar Problems

Question 1: Expand [√2(cos pi/4+i sin pi/4]10.

Solution:

As per DeMoivre’s theorem: (cosθ + sinθ)n = cos(nθ) + i sin(nθ).

[\sqrt{2}(cos(\frac{π}{4})+ i sin(\frac{π}{4})]^{10} = (\sqrt2)^{10}[cos(\frac{10π}{4})+i\ sin(\frac{10π}{4})]

= 32(0 + (-i)

= -32i

Hence, [√2(cos pi/4 + i sin pi/4)]10 = -32i.

Question 2: Expand [√2(cos π/4 +i sin π/4)]5.

Solution:

As per DeMoivre’s theorem: (cosθ + sinθ)n = cos(nθ) + i sin(nθ).

 [\sqrt{2}cos(\frac{π}{4})+i\ sin(\frac{π}{4})]^5 = (\sqrt{2})^{5}[cos(\frac{5\pi}{4})+i\ sin(\frac{5\pi}{4})]

= -4 – 4i

Hence, [√2(cos π/4 +i sin π/4)]5 = -4 – 4i.

Question 3. Expand [2\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^6  .

Solution:

As per DeMoivre’s theorem: (cosθ + sinθ)n = cos(nθ) + i sin(nθ).

[2\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^6 = (2\sqrt{2})^{6}[cos(\frac{6\pi}{4})+i\ sin(\frac{6\pi}{4})]

= 512 (-i)

= -512i

Hence, [2\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^6   = -512i.

Question 4. Expand [\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^{18}

Solution:

As per DeMoivre’s theorem: (cosθ + sinθ)n = cos(nθ) + i sin(nθ).

[\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^{18} = (\sqrt{2})^{18}[cos(\frac{18\pi}{4})+i\ sin(\frac{18\pi}{4})]

= 512i
Hence, [\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^{18}   = 512i.

Question 5. Expand [2\sqrt{3}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^{31}  .

Solution:

As per DeMoivre’s theorem: (cosθ + sinθ)n = cos(nθ) + i sin(nθ).

[2\sqrt{3}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^{31} = (\sqrt{2})^{31}[cos(\frac{31\pi}{4})+i\ sin(\frac{31\pi}{4})]  .


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!