Question:
If
\[
(x-iy)^{1/3}=a-ib,
\]
then the value of
\[
\frac{x}{2a}+\frac{y}{2b}
\]
is:
If
\[
(x-iy)^{1/3}=a-ib,
\]
then the value of
\[
\frac{x}{2a}+\frac{y}{2b}
\]
is:
Show Hint
For complex number problems involving powers, first expand using algebraic identities and then compare real and imaginary parts separately.
Updated On: Jun 24, 2026
- \(2(a^2-b^2)\)
- \(4(a^2-b^2)\)
- \(a^2-b^2\)
- \(\dfrac{1}{2}(a^2-b^2)\)
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The Correct Option is A
Solution and Explanation
Step 1: Cube both sides.
Given, \[ (x-iy)^{1/3}=a-ib \] Cubing both sides, \[ x-iy=(a-ib)^3 \]
Step 2: Expand \((a-ib)^3\).
Using \[ (A-B)^3=A^3-3A^2B+3AB^2-B^3, \] we get \[ (a-ib)^3 =a^3-3a^2(ib)+3a(ib)^2-(ib)^3 \] Now, \[ (ib)^2=i^2b^2=-b^2 \] and \[ (ib)^3=i^3b^3=-ib^3 \] Therefore, \[ (a-ib)^3 =a^3-3a^2ib-3ab^2+ib^3 \] Grouping real and imaginary parts, \[ (a-ib)^3 =(a^3-3ab^2)-i(3a^2b-b^3) \] Thus, \[ x=a^3-3ab^2 \] and \[ y=3a^2b-b^3 \]
Step 3: Compute \[ \frac{x}{2a}+\frac{y}{2b} \] Substituting the values of \(x\) and \(y\), \[ \frac{x}{2a} = \frac{a^3-3ab^2}{2a} = \frac{a^2-3b^2}{2} \] Similarly, \[ \frac{y}{2b} = \frac{3a^2b-b^3}{2b} = \frac{3a^2-b^2}{2} \] Adding, \[ \frac{x}{2a}+\frac{y}{2b} = \frac{a^2-3b^2+3a^2-b^2}{2} \] \[ = \frac{4a^2-4b^2}{2} \] \[ =2(a^2-b^2) \]
Step 4: Final conclusion.
Hence, \[ \boxed{2(a^2-b^2)} \]
Given, \[ (x-iy)^{1/3}=a-ib \] Cubing both sides, \[ x-iy=(a-ib)^3 \]
Step 2: Expand \((a-ib)^3\).
Using \[ (A-B)^3=A^3-3A^2B+3AB^2-B^3, \] we get \[ (a-ib)^3 =a^3-3a^2(ib)+3a(ib)^2-(ib)^3 \] Now, \[ (ib)^2=i^2b^2=-b^2 \] and \[ (ib)^3=i^3b^3=-ib^3 \] Therefore, \[ (a-ib)^3 =a^3-3a^2ib-3ab^2+ib^3 \] Grouping real and imaginary parts, \[ (a-ib)^3 =(a^3-3ab^2)-i(3a^2b-b^3) \] Thus, \[ x=a^3-3ab^2 \] and \[ y=3a^2b-b^3 \]
Step 3: Compute \[ \frac{x}{2a}+\frac{y}{2b} \] Substituting the values of \(x\) and \(y\), \[ \frac{x}{2a} = \frac{a^3-3ab^2}{2a} = \frac{a^2-3b^2}{2} \] Similarly, \[ \frac{y}{2b} = \frac{3a^2b-b^3}{2b} = \frac{3a^2-b^2}{2} \] Adding, \[ \frac{x}{2a}+\frac{y}{2b} = \frac{a^2-3b^2+3a^2-b^2}{2} \] \[ = \frac{4a^2-4b^2}{2} \] \[ =2(a^2-b^2) \]
Step 4: Final conclusion.
Hence, \[ \boxed{2(a^2-b^2)} \]
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