Question

Difficulty: MediumComplex Numbers and Operations

For the complex number z=3+5i2iz = \frac{3 + 5i}{2 - i}, where i=1i = \sqrt{-1}, what is the real part of zz?

  1. 15\frac{1}{5}Answer
  2. B
    115\frac{11}{5}
  3. C
    32\frac{3}{2}
  4. D
    13\frac{1}{3}
  5. E
    15-\frac{1}{5}

Answer

The real part of the complex number is 15\frac{1}{5}.
To find the real part of the complex fraction, we multiply both the numerator and the denominator by the complex conjugate of the denominator, 2+i2 + i. This yields a denominator of (2i)(2+i)=4i2=5(2-i)(2+i) = 4 - i^2 = 5. The numerator becomes (3+5i)(2+i)=6+3i+10i+5i2=1+13i(3+5i)(2+i) = 6 + 3i + 10i + 5i^2 = 1 + 13i. Writing the result in standard form gives 15+135i\frac{1}{5} + \frac{13}{5}i, so the real part is 15\frac{1}{5}.

Step-by-Step Solution

1
Identify the complex conjugate of the denominator, which is 2+i2 + i, and set up the multiplication for both the numerator and the denominator of zz.
(3+5i)(2+i)(2i)(2+i)\frac{(3 + 5i)(2 + i)}{(2 - i)(2 + i)}
Multiplying the numerator and denominator by the conjugate of the denominator allows us to simplify the expression and eliminate the imaginary unit from the denominator.
2
Expand the numerator using binomial multiplication and simplify using the identity i2=1i^2 = -1.
6+3i+10i+5i2=6+13i5=1+13i6 + 3i + 10i + 5i^2 = 6 + 13i - 5 = 1 + 13i
This determines the simplified complex expression in the numerator.
3
Expand the denominator and simplify using the identity i2=1i^2 = -1.
(2i)(2+i)=4i2=4(1)=5(2 - i)(2 + i) = 4 - i^2 = 4 - (-1) = 5
This simplifies the denominator into a real number.
4
Combine the simplified numerator and denominator to write zz in standard form a+bia + bi and identify the real part aa.
z=1+13i5=15+135iz = \frac{1 + 13i}{5} = \frac{1}{5} + \frac{13}{5}i, so the real part is 15\frac{1}{5}.
Expressing the complex number in standard form separates the real and imaginary components.

Key Concept

Complex Division and Simplification
Estimated Time:1m 30s
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