Question:
If \( a \) and \( b \) are real numbers and \( (a + ib)^{11} = 1 + 3i \), then \( (b + ia)^{11} \) is equal to
If \( a \) and \( b \) are real numbers and \( (a + ib)^{11} = 1 + 3i \), then \( (b + ia)^{11} \) is equal to
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Use conjugate and factorization tricks when swapping real and imaginary parts.
Updated On: May 1, 2026
- \( i + 3 \)
- \( 1 + 3i \)
- \( 1 - 3i \)
- \( 0 \)
- \( -i - 3 \)
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The Correct Option is A
Solution and Explanation
Concept:
Observe:
\[
b + ia = i(a - ib)
\]
Also:
\[
(a - ib) = \overline{(a+ib)}
\]
Step 1: Rewrite expression.
\[ (b + ia)^{11} = [i(a - ib)]^{11} = i^{11}(a - ib)^{11} \]
Step 2: Use conjugate property.
\[ (a - ib)^{11} = \overline{(a+ib)^{11}} = \overline{(1+3i)} = 1 - 3i \]
Step 3: Compute \( i^{11} \).
\[ 11 \mod 4 = 3 \Rightarrow i^{11} = i^3 = -i \]
Step 4: Multiply.
\[ (-i)(1 - 3i) = -i + 3i^2 = -i -3 \]
Step 5: Rearrange.
\[ = -(3 + i) = i + 3 \]
Step 1: Rewrite expression.
\[ (b + ia)^{11} = [i(a - ib)]^{11} = i^{11}(a - ib)^{11} \]
Step 2: Use conjugate property.
\[ (a - ib)^{11} = \overline{(a+ib)^{11}} = \overline{(1+3i)} = 1 - 3i \]
Step 3: Compute \( i^{11} \).
\[ 11 \mod 4 = 3 \Rightarrow i^{11} = i^3 = -i \]
Step 4: Multiply.
\[ (-i)(1 - 3i) = -i + 3i^2 = -i -3 \]
Step 5: Rearrange.
\[ = -(3 + i) = i + 3 \]
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