Compute (5 - 3i)³.
A125 + 27i
B125 - 27i
C-10 - 198i
D-10 + 198i
Answer & Solution
Correct answer: C. -10 - 198i
(a - b)³ = a³ - 3a²b + 3ab² - b³. Here a = 5, b = 3i. a³ = 125. 3a²b = 3·25·3i = 225i. 3ab² = 3·5·(3i)² = 15·(-9) = -135. b³ = (3i)³ = 27i³ = -27i. Sum: 125 - 225i + (-135) - (-27i) = (125 - 135) + (-225 + 27)i = -10 - 198i.
Related questions
Using De Moivre's theorem, the complex roots of $z^3=1$ are:If $\alpha$ and $\beta$ are roots of $z^2+4z+8=0$, then $\dfrac{\alpha+\beta+4i}{\alpha\beThe locus $\left|\dfrac{z-2}{z+3i}\right|=4$ simplifies to which equation?Simplify i raised to the power 27. What is the result?Multiply (2 + 3i) by (4 - i). What is the product in standard form?How is the complex conjugate of a complex number obtained?Add the complex numbers 5 - 2i and 3 + 7i. What is the sum?In the complex number 7 + 4i, which part is the real part?