Find The Indicated Power Using De Moivre's Theorem

Find the Indicated Power Using De Moivre’s Theorem
De Moivre’s theorem provides a straightforward way to raise complex numbers to integer powers by converting them to polar form. By applying the theorem, you can compute powers quickly, avoid lengthy algebraic expansion, and gain insight into the geometric interpretation of complex exponentiation. This guide walks you through the concept, the step‑by‑step procedure, worked examples, and common pitfalls so you can confidently find any indicated power using De Moivre’s theorem.

Understanding De Moivre’s Theorem

For any real number θ and integer n, De Moivre’s theorem states

[ (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta). ]

When a complex number (z) is expressed in polar form

[ z = r(\cos\theta + i\sin\theta) = re^{i\theta}, ]

raising (z) to the (n^{\text{th}}) power becomes

[ z^n = \bigl[r(\cos\theta + i\sin\theta)\bigr]^n = r^n\bigl(\cos(n\theta) + i\sin(n\theta)\bigr). ]

Thus, the modulus is raised to the power (n) and the argument is multiplied by (n). This compact formula is the core of finding indicated powers efficiently.

Steps to Find the Indicated Power

Follow these systematic steps whenever you need to compute (z^n) for a given complex number (z) and integer exponent (n):

1. Convert to Polar Form

   • Compute the modulus (r = \sqrt{a^2 + b^2}) for (z = a + bi).
   • Determine the argument (\theta = \operatorname{atan2}(b, a)) (principal value in ((-\pi, \pi]) or ([0, 2\pi)) as required).
   • Write (z = r(\cos\theta + i\sin\theta)).

2. Apply De Moivre’s Theorem

   • Raise the modulus: (r^n).
   • Multiply the argument: (n\theta).
   • Form the polar result: (z^n = r^n\bigl(\cos(n\theta) + i\sin(n\theta)\bigr)).

3. Convert Back to Rectangular Form (if needed)

   • Use (a' = r^n\cos(n\theta)) and (b' = r^n\sin(n\theta)) to obtain (z^n = a' + b'i).
   • Simplify trigonometric values when (\theta) corresponds to standard angles (e.g., (0, \pi/6, \pi/4, \pi/3, \pi/2)).

4. Check for Periodicity (optional)

   • Because adding (2\pi) to (\theta) does not change the complex number, you may reduce (n\theta) modulo (2\pi) to keep the argument within a preferred interval.

These steps work for positive, negative, and zero exponents. For negative (n), treat (z^{-n} = 1/(z^n)) or directly apply the theorem with a negative multiplier to the argument Easy to understand, harder to ignore. That alone is useful..

Example Problems

Example 1: Positive Integer Power

Find ((1 + i)^4) Not complicated — just consistent..

Solution

1. Modulus: (r = \sqrt{1^2 + 1^2} = \sqrt{2}).
   Argument: (\theta = \tan^{-1}(1/1) = \pi/4).
   Polar form: (1+i = \sqrt{2}\bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\bigr)).

2. Apply theorem with (n=4):
   [ (1+i)^4 = (\sqrt{2})^4\bigl(\cos(4\cdot\frac{\pi}{4}) + i\sin(4\cdot\frac{\pi}{4})\bigr) = 4\bigl(\cos\pi + i\sin\pi\bigr). ]

3. Evaluate trigonometric functions: (\cos\pi = -1,\ \sin\pi = 0).
   Hence ((1+i)^4 = 4(-1 + 0i) = -4) Which is the point..

Example 2: Negative Integer Power

Compute ((2-2i)^{-3}).

Solution

1. Modulus: (r = \sqrt{2^2 + (-2)^2} = \sqrt{8}=2\sqrt{2}).
   Argument: (\theta = \tan^{-1}\bigl(\frac{-2}{2}\bigr) = -\pi/4) (or (7\pi/4)).
   Polar form: (2-2i = 2\sqrt{2}\bigl(\cos(-\pi/4) + i\sin(-\pi/4)\bigr)).

2. Apply theorem with (n=-3):
   [ (2-2i)^{-3} = (2\sqrt{2})^{-3}\bigl(\cos(-3\cdot-\pi/4) + i\sin(-3\cdot-\pi/4)\bigr) = \frac{1}{(2\sqrt{2})^3}\bigl(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\bigr). ]

   ((2\sqrt{2})^3 = 8\cdot2\sqrt{2}=16\sqrt{2}). So the modulus factor is (1/(16\sqrt{2})).

3. Trigonometric values: (\cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2},\ \sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2}).
   Multiply:
   [ \frac{1}{16\sqrt{2}}\left(-\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\right) = \frac{1}{16}\left(-\frac{1}{2} + i\frac{1}{2}\right) = -\frac{1}{32} + i\frac{1}{32}. ]

Thus ((2-2i)^{-3} = -\dfrac{1}{32} + \dfrac{1}{32}i) That's the part that actually makes a difference..

Example 3: Power of a Pure Imaginary Number

Find ((3i)^5).

Solution

1. Write (3i = 0 + 3i). Modulus: (r = 3). Argument: (\theta = \pi/2) (point on positive imaginary axis).
   Polar: (3i = 3\bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\bigr)) Simple as that..

2. Apply theorem with (n=5):
   [ (3i)^5 = 3^5\bigl(\cos(5\cdot\frac{\pi}{2}) + i\sin(5\cdot\frac{\pi}{2})\bigr) = 243\bigl(\cos\frac{5\pi}{2} + i\sin\frac{5\pi}{2}\bigr). ]

3. Reduce angle: (\frac{5\pi}{2} = 2\pi + \frac{\pi}{2}). Cosine and sine are periodic with