Question

Difficulty: MediumComplex Numbers and Operations

For the imaginary unit ii, where i2=1i^2 = -1, the complex number zz is defined by z=11+3i3iz = \frac{11 + 3i}{3 - i}. What is the real part of zz?

Answer: 3

Answer

The real part of the complex number is 3.
By multiplying both the numerator and the denominator of 11+3i3i\frac{11 + 3i}{3 - i} by the conjugate of the denominator, 3+i3 + i, we obtain (11+3i)(3+i)(3i)(3+i)=33+11i+9i+3i29i2=30+20i10=3+2i\frac{(11+3i)(3+i)}{(3-i)(3+i)} = \frac{33 + 11i + 9i + 3i^2}{9 - i^2} = \frac{30 + 20i}{10} = 3 + 2i. The real part of this complex number is the term without ii, which is 3.

Step-by-Step Solution

1
Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator, which is 3+i3 + i.
z=(11+3i)(3+i)(3i)(3+i)z = \frac{(11 + 3i)(3 + i)}{(3 - i)(3 + i)}
To eliminate the imaginary unit from the denominator.
2
Expand and simplify the numerator using the distributive property and substituting 1-1 for i2i^2.
(11+3i)(3+i)=33+11i+9i+3i2=33+20i+3(1)=30+20i(11 + 3i)(3 + i) = 33 + 11i + 9i + 3i^2 = 33 + 20i + 3(-1) = 30 + 20i
To combine the real and imaginary terms of the numerator.
3
Expand and simplify the denominator using the difference of squares property and substituting 1-1 for i2i^2.
(3i)(3+i)=9i2=9(1)=10(3 - i)(3 + i) = 9 - i^2 = 9 - (-1) = 10
To find the real number denominator.
4
Divide each term in the simplified numerator by the denominator.
z=30+20i10=3+2iz = \frac{30 + 20i}{10} = 3 + 2i
To express the complex number in the standard form a+bia + bi.
5
Extract the real part of the resulting complex number 3+2i3 + 2i.
3
The real part of a complex number in the form a+bia + bi is aa.

Key Concept

Division of complex numbers using the complex conjugate

Alternative Method

Instead of simplifying the fraction directly, assume the resulting complex number is x+yix + yi, where xx represents the real part and yy represents the imaginary part. We can set up the equation x+yi=11+3i3ix + yi = \frac{11 + 3i}{3 - i} and multiply both sides by 3i3 - i to get (x+yi)(3i)=11+3i(x + yi)(3 - i) = 11 + 3i. Expanding the left side gives (3x+y)+(3yx)i=11+3i(3x + y) + (3y - x)i = 11 + 3i. Equating the real and imaginary parts yields a system of linear equations: 3x+y=113x + y = 11 and x+3y=3-x + 3y = 3. Multiplying the second equation by 3 and adding it to the first equation gives 10y=2010y = 20, which means y=2y = 2. Substituting y=2y = 2 back into the first equation yields 3x+2=113x + 2 = 11, which simplifies to 3x=93x = 9, or x=3x = 3. The real part is therefore 3.
Estimated Time:1m 30s
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