Question:
If
\[
z=\frac{(1-i)^3}{(\sqrt{3}-i)^2},
\]
then the complex conjugate of \(z\) is:
If
\[
z=\frac{(1-i)^3}{(\sqrt{3}-i)^2},
\]
then the complex conjugate of \(z\) is:
Show Hint
Whenever powers and quotients of complex numbers appear, convert them into polar form. De Moivre's theorem significantly reduces lengthy algebraic calculations.
Updated On: Jun 10, 2026
- \(\dfrac{\sqrt{3}+1}{4}+\dfrac{\sqrt{3}-1}{4}i\)
- \(\dfrac{\sqrt{3}-1}{4}+\dfrac{\sqrt{3}+1}{4}i\)
- \(\dfrac{\sqrt{3}+1}{4}-\dfrac{\sqrt{3}-1}{4}i\)
- \(\dfrac{\sqrt{3}-1}{4}-\dfrac{\sqrt{3}+1}{4}i\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
To determine the complex conjugate of a complex number involving powers and quotients, it is often convenient to convert the numbers into polar form. In polar form, multiplication and division become much simpler because moduli divide and arguments subtract.
Recall that if
\[
z=r(\cos\theta+i\sin\theta),
\]
then
\[
\overline{z}=r(\cos\theta-i\sin\theta).
\]
Step 1: Simplify the numerator Consider \[ (1-i)^3. \] First write \(1-i\) in polar form. Its modulus is \[ |1-i|=\sqrt{1^2+(-1)^2} =\sqrt2. \] Its argument is \[ -\frac{\pi}{4}. \] Therefore, \[ 1-i=\sqrt2 \left( \cos\frac{-\pi}{4} +i\sin\frac{-\pi}{4} \right). \] Hence, \[ (1-i)^3 = (\sqrt2)^3 \left( \cos\frac{-3\pi}{4} +i\sin\frac{-3\pi}{4} \right). \] \[ = 2\sqrt2 \left( \cos\frac{-3\pi}{4} +i\sin\frac{-3\pi}{4} \right). \]
Step 2: Simplify the denominator Consider \[ (\sqrt3-i)^2. \] The modulus of \(\sqrt3-i\) is \[ \sqrt{3+1}=2. \] Its argument is \[ -\frac{\pi}{6}. \] Therefore, \[ (\sqrt3-i)^2 = 4 \left( \cos\frac{-\pi}{3} +i\sin\frac{-\pi}{3} \right). \]
Step 3: Divide the two complex numbers Hence \[ z = \frac{2\sqrt2}{4} \left[ \cos\left( -\frac{3\pi}{4}+\frac{\pi}{3} \right) +i\sin\left( -\frac{3\pi}{4}+\frac{\pi}{3} \right) \right]. \] \[ = \frac{\sqrt2}{2} \left[ \cos\left(-\frac{5\pi}{12}\right) +i\sin\left(-\frac{5\pi}{12}\right) \right]. \]
Step 4: Find the conjugate The conjugate is \[ \overline z = \frac{\sqrt2}{2} \left[ \cos\left(\frac{5\pi}{12}\right) +i\sin\left(\frac{5\pi}{12}\right) \right]. \] Using \[ \cos75^\circ = \frac{\sqrt3-1}{2\sqrt2}, \] and \[ \sin75^\circ = \frac{\sqrt3+1}{2\sqrt2}, \] we obtain \[ \overline z = \frac{\sqrt3+1}{4} -\frac{\sqrt3-1}{4}i. \] Therefore, \[ \boxed{\overline z = \frac{\sqrt3+1}{4} -\frac{\sqrt3-1}{4}i} \]
Step 1: Simplify the numerator Consider \[ (1-i)^3. \] First write \(1-i\) in polar form. Its modulus is \[ |1-i|=\sqrt{1^2+(-1)^2} =\sqrt2. \] Its argument is \[ -\frac{\pi}{4}. \] Therefore, \[ 1-i=\sqrt2 \left( \cos\frac{-\pi}{4} +i\sin\frac{-\pi}{4} \right). \] Hence, \[ (1-i)^3 = (\sqrt2)^3 \left( \cos\frac{-3\pi}{4} +i\sin\frac{-3\pi}{4} \right). \] \[ = 2\sqrt2 \left( \cos\frac{-3\pi}{4} +i\sin\frac{-3\pi}{4} \right). \]
Step 2: Simplify the denominator Consider \[ (\sqrt3-i)^2. \] The modulus of \(\sqrt3-i\) is \[ \sqrt{3+1}=2. \] Its argument is \[ -\frac{\pi}{6}. \] Therefore, \[ (\sqrt3-i)^2 = 4 \left( \cos\frac{-\pi}{3} +i\sin\frac{-\pi}{3} \right). \]
Step 3: Divide the two complex numbers Hence \[ z = \frac{2\sqrt2}{4} \left[ \cos\left( -\frac{3\pi}{4}+\frac{\pi}{3} \right) +i\sin\left( -\frac{3\pi}{4}+\frac{\pi}{3} \right) \right]. \] \[ = \frac{\sqrt2}{2} \left[ \cos\left(-\frac{5\pi}{12}\right) +i\sin\left(-\frac{5\pi}{12}\right) \right]. \]
Step 4: Find the conjugate The conjugate is \[ \overline z = \frac{\sqrt2}{2} \left[ \cos\left(\frac{5\pi}{12}\right) +i\sin\left(\frac{5\pi}{12}\right) \right]. \] Using \[ \cos75^\circ = \frac{\sqrt3-1}{2\sqrt2}, \] and \[ \sin75^\circ = \frac{\sqrt3+1}{2\sqrt2}, \] we obtain \[ \overline z = \frac{\sqrt3+1}{4} -\frac{\sqrt3-1}{4}i. \] Therefore, \[ \boxed{\overline z = \frac{\sqrt3+1}{4} -\frac{\sqrt3-1}{4}i} \]
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