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Question

Difficulty: MediumComplex Numbers and Operations

For the imaginary unit ii, where i2=−1i^2 = -1, what is the real part of the complex number z=(2−i)3+9iz = (2 - i)^3 + 9i?

Answer: 2

Answer

The real part of the complex number is 2.
Expanding the expression (2−i)3(2 - i)^3 yields 2−11i2 - 11i. Adding 9i9i gives 2−2i2 - 2i. The real part of this complex number is the term without ii, which is 2.

Step-by-Step Solution

1
Expand the squared binomial (2−i)2(2 - i)^2.
3−4i3 - 4i
Apply the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 and substitute i2=−1i^2 = -1.
2
Multiply the result of the square by (2−i)(2 - i) to calculate (2−i)3(2 - i)^3.
2−11i2 - 11i
Distribute the terms (3−4i)(2−i)=6−3i−8i+4i2(3 - 4i)(2 - i) = 6 - 3i - 8i + 4i^2 and substitute i2=−1i^2 = -1.
3
Add 9i9i to the simplified cube to find the complex number zz.
2−2i2 - 2i
Combine the imaginary components: −11i+9i=−2i-11i + 9i = -2i.
4
Identify the real part of zz.
2
The real part of a complex number a+bia + bi is aa.

Key Concept

Expanding complex binomials and simplifying powers of the imaginary unit
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Topics

Complex Numbers and Operations

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