Question

Difficulty: Very hardComplex Numbers and Operations

For the imaginary unit ii, where i2=1i^2 = -1, the complex number zz is defined as z=(1+2i)32iz = \frac{(1 + 2i)^3}{2 - i}. What is the absolute value of zz?

Answer: 5

Answer

5
The complex number zz simplifies to 43i-4 - 3i. The absolute value of a complex number a+bia + bi is defined as a2+b2\sqrt{a^2 + b^2}, which gives (4)2+(3)2=16+9=5\sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = 5. Alternatively, using properties of the modulus, z=1+2i32i=(12+22)322+(1)2=(5)35=5|z| = \frac{|1 + 2i|^3}{|2 - i|} = \frac{(\sqrt{1^2 + 2^2})^3}{\sqrt{2^2 + (-1)^2}} = \frac{(\sqrt{5})^3}{\sqrt{5}} = 5.

Step-by-Step Solution

1
Expand the numerator (1+2i)3(1 + 2i)^3
-11 - 2i
To begin simplifying the complex fraction by expanding the cubed binomial in the numerator.
2
Divide the result by the denominator 2i2 - i by multiplying both numerator and denominator by the complex conjugate 2+i2 + i
-4 - 3i
To express the quotient in standard complex form a+bia + bi.
3
Find the absolute value of the simplified complex number z=43iz = -4 - 3i
5
To calculate the final magnitude of the complex number using a+bi=a2+b2|a + bi| = \sqrt{a^2 + b^2}.

Key Concept

Complex number operations, complex division, and absolute value (modulus) of complex numbers
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