Compute the value of (1 + i)^10 using De Moivre's theorem.
Answer and explanation
Correct answer: A. 32*i
- Step 1: Express 1 + i in polar form: r = sqrt(1^2 + 1^2) = sqrt(2), theta = pi/4.
- Step 2: Apply De Moivre's theorem: (1 + i)^10 = (sqrt(2))^10 (cos(10pi/4) + isin(10pi/4)).
- Step 3: (sqrt(2))^10 = 2^5 = 32.
- Step 4: 10pi/4 = 5pi/2 = 2*pi + pi/2, so cos(pi/2) = 0 and sin(pi/2) = 1.
- Step 5: Result = 32 (0 + i) = 32i.
Answer
32*i.
Why not the others: -32i has the wrong sign for sin(5pi/2); 32 assumes argument 0; -32 assumes argument pi.
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