Question:
The modulus of the product of all the values of \( (2+3i)^{3/5} \) is:
The modulus of the product of all the values of \( (2+3i)^{3/5} \) is:
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For \( z^{p/q} \), the modulus of the product of all \( q \) roots is simply \( |z|^p \).
Updated On: Jun 9, 2026
- \( \sqrt{2197} \)
- \( \sqrt{2245} \)
- \( \sqrt{135} \)
- \( \sqrt{489} \)
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The Correct Option is A
Solution and Explanation
Concept:
If \( z = r(\cos \theta + i\sin \theta) \), then the \( n \) roots of \( z^{p/q} \) are \( r^{p/q}(\cos \phi_k + i\sin \phi_k) \). The modulus of each root is \( r^{p/q} \).
Step 1: Find the modulus of the base \( z \).
\( |z| = \sqrt{2^2 + 3^2} = \sqrt{4+9} = \sqrt{13} \).
Step 2: Find the product modulus.
There are 5 roots. Each root has modulus \( (\sqrt{13})^{3/5} \). The product of 5 roots has modulus equal to \( ( (\sqrt{13})^{3/5} )^5 = (\sqrt{13})^3 \).
Step 3: Simplify \( (\sqrt{13})^3 \).
\( \sqrt{13} \cdot \sqrt{13} \cdot \sqrt{13} = 13\sqrt{13} = \sqrt{169 \cdot 13} = \sqrt{2197} \). 2197
Step 1: Find the modulus of the base \( z \).
\( |z| = \sqrt{2^2 + 3^2} = \sqrt{4+9} = \sqrt{13} \).
Step 2: Find the product modulus.
There are 5 roots. Each root has modulus \( (\sqrt{13})^{3/5} \). The product of 5 roots has modulus equal to \( ( (\sqrt{13})^{3/5} )^5 = (\sqrt{13})^3 \).
Step 3: Simplify \( (\sqrt{13})^3 \).
\( \sqrt{13} \cdot \sqrt{13} \cdot \sqrt{13} = 13\sqrt{13} = \sqrt{169 \cdot 13} = \sqrt{2197} \). 2197
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