Express (1+i)4(1 + i)^4 using De Moivre's theorem.

Answer: (1+i)4=−4(1 + i)^4 = -4

Step-by-step solution

Step 1: Find the modulus and argument of 1+i1 + i

To apply De Moivre's theorem, we first represent the complex number z=1+iz = 1 + i in polar form r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta). The modulus is r=12+12=2r = \sqrt{1^2 + 1^2} = \sqrt{2}. Since the point (1,1)(1, 1) lies in the first quadrant, the principal argument is θ=π4\theta = \frac{\pi}{4}.

Step 2: Write 1+i1 + i in polar form

Substituting the modulus r=2r = \sqrt{2} and argument θ=π4\theta = \frac{\pi}{4} gives the polar form 1+i=2(cos⁡π4+isin⁡π4)1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right).

Step 3: Apply De Moivre's theorem

De Moivre's theorem states that (cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ)(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta) for any integer nn. Raising both sides to the power 44, we raise the modulus 2\sqrt{2} to the power 44, giving (2)4=4(\sqrt{2})^4 = 4, and multiply the angle π4\frac{\pi}{4} by 44.

Step 4: Evaluate the trigonometric values to find the final result

Evaluating the trigonometric functions at π\pi, we have cos⁡π=−1\cos\pi = -1 and sin⁡π=0\sin\pi = 0. Multiplying by 44 gives 4(−1+0i)=−44(-1 + 0i) = -4. In standard a+bia + bi form, this is −4+0i-4 + 0i.

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