Examkin

Question

Difficulty: HardComplex Numbers and Operations

For the imaginary unit ii, if the complex number zz is defined by z=(1+2i)22−iz = \frac{(1 + 2i)^2}{2 - i}, what is the real part of zz?

  1. −2-2Answer
  2. B
    −65-\frac{6}{5}
  3. C
    65\frac{6}{5}
  4. D
    −103-\frac{10}{3}
  5. E
    −32-\frac{3}{2}

Answer

The real part of the complex number zz is −2-2.
To find the real part of the complex number, we first simplify the expression by expanding the squared binomial in the numerator, which yields −3+4i-3 + 4i. Next, we rationalize the fraction by multiplying both the numerator and the denominator by the complex conjugate of the denominator, 2+i2 + i. This multiplication yields −10+5i5\frac{-10 + 5i}{5}. Dividing both the real and imaginary terms by 55 results in the standard form −2+i-2 + i. Thus, the real part of this complex number is −2-2.

Step-by-Step Solution

1
Expand the squared binomial in the numerator of the expression for zz.
(1+2i)2=12+2(1)(2i)+(2i)2=1+4i+4i2(1 + 2i)^2 = 1^2 + 2(1)(2i) + (2i)^2 = 1 + 4i + 4i^2
Apply the algebraic identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2.
2
Simplify the expanded numerator using the definition of the imaginary unit.
1+4i+4(−1)=−3+4i1 + 4i + 4(-1) = -3 + 4i
Since i2=−1i^2 = -1, the term 4i24i^2 simplifies to −4-4. Combining this with 11 gives the real part −3-3.
3
Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator to rationalize it.
z=−3+4i2−i⋅2+i2+i=(−3+4i)(2+i)(2−i)(2+i)z = \frac{-3 + 4i}{2 - i} \cdot \frac{2 + i}{2 + i} = \frac{(-3 + 4i)(2 + i)}{(2 - i)(2 + i)}
Multiplying the denominator by its complex conjugate, 2+i2 + i, eliminates the imaginary unit from the denominator.
4
Expand and simplify the numerator and denominator.
z=−6−3i+8i+4i24−i2=−6+5i−44−(−1)=−10+5i5z = \frac{-6 - 3i + 8i + 4i^2}{4 - i^2} = \frac{-6 + 5i - 4}{4 - (-1)} = \frac{-10 + 5i}{5}
Using the distributive property in the numerator gives −6+5i+4i2-6 + 5i + 4i^2. Since i2=−1i^2 = -1, this simplifies to −10+5i-10 + 5i. In the denominator, (2−i)(2+i)=4−i2=5(2-i)(2+i) = 4 - i^2 = 5.
5
Divide both terms of the simplified numerator by the denominator to express zz in standard form a+bia + bi.
z=−2+iz = -2 + i
Dividing the real part −10-10 by 55 yields the real part −2-2, and dividing the imaginary part 5i5i by 55 yields the imaginary part ii.

Key Concept

Division of complex numbers using the complex conjugate
Estimated Time:2m 0s
Rate this question

Topics

Complex Numbers and Operations

More questions from this topic