Question

If $\frac{(1+i)(2+3i)(3-4i)}{(2-3i)(1-i)(3+4i)} = a + ib$, then $a^2 + b^2 =$

• A. $132$
• B. $25$
• C. $144$
• D. $128$
• E. $1$

Hint:

Don't waste time multiplying out complex fractions if you only need the modulus ($a^2 + b^2$). If every term on top has a corresponding conjugate on the bottom, the modulus is guaranteed to be 1.

Answer 1

Answer: (E)

Concept: The value $a^2 + b^2$ is the square of the modulus of the complex number $z = a + ib$. We use the property that the modulus of a product/quotient is the product/quotient of the moduli: $| \frac{z_1 z_2 z_3}{z_4 z_5 z_6} | = \frac{|z_1| |z_2| |z_3|}{|z_4| |z_5| |z_6|}$.

Step 1: Identify conjugate pairs in the numerator and denominator.
Look at the terms in the expression:
• Numerator: $(1+i)$, $(2+3i)$, $(3-4i)$
• Denominator: $(1-i)$, $(2-3i)$, $(3+4i)$ Each term in the numerator has its complex conjugate in the denominator.

Step 2: Apply the property of moduli.
We know that $|z| = |\bar{z}|$. Therefore:
• $|1+i| = |1-i| = \sqrt{1^2 + 1^2} = \sqrt{2}$
• $|2+3i| = |2-3i| = \sqrt{2^2 + 3^2} = \sqrt{13}$
• $|3-4i| = |3+4i| = \sqrt{3^2 + (-4)^2} = 5$

Step 3: Calculate $|z|^2$.
\[ |z| = \frac{|1+i| |2+3i| |3-4i|}{|2-3i| |1-i| |3+4i|} = \frac{\sqrt{2} \cdot \sqrt{13} \cdot 5}{\sqrt{13} \cdot \sqrt{2} \cdot 5} = 1 \] Since $|z| = 1$, then $a^2 + b^2 = |z|^2 = 1^2 = 1$.