Complex Numbers
Rotation of complex numbers
Grade 11
Question:
<p>A complex number <em>z</em> is rotated in anticlockwise direction by an angle <em>α</em> and we get <em>z'</em> and if the same complex number <em>z</em> is rotated by an angle <em>α</em> in clockwise direction and we get <em>z''</em>, then</p>
<p>(1) \(z', z, z''\) are in G.P.</p>
<p>(2) \(z', z, z''\) are in H.P.</p>
<p>(3) \(z' + z'' = 2z \cos\alpha\)</p>
<p>(4) \(z'^2 + z''^2 = 2z^2 \cos 2\alpha\)</p>
Step-by-Step Solution
Key Concept: Rotation of a complex number z by angle α is achieved by multiplying z by e^(iα); anticlockwise rotation uses +α while clockwise uses -α, creating conjugate relationships.
<p><strong>Step 1:</strong> Define rotations. If z is rotated anticlockwise by angle α: <br/>z' = z·e^(iα) = z(cos α + i sin α)</p><p><strong>Step 2:</strong> If z is rotated clockwise by angle α:<br/>z'' = z·e^(-iα) = z(cos α - i sin α)</p><p><strong>Step 3:</strong> Compare properties:<br/>• z' + z'' = z·e^(iα) + z·e^(-iα) = z(2cos α) = 2(Re(z') + Re(z''))/2 = Real if z real<br/>• z'·z'' = z·e^(iα)·z·e^(-iα) = z²·e^0 = z² (real when z purely real or imaginary)<br/>• |z'| = |z·e^(iα)| = |z| and |z''| = |z·e^(-iα)| = |z|, so |z'| = |z''|<br/>• arg(z') = arg(z) + α and arg(z'') = arg(z) - α, so arg(z') - arg(z'') = 2α</p><p><strong>Step 4:</strong> Verify standard results:<br/>A) |z'| = |z''| = |z| ✓<br/>C) arg(z') - arg(z'') = 2α ✓<br/>D) z' + z'' = 2|z|cos(arg(z) + α) ✓</p><p>∴ Answer: A, C, D</p>
Correct Answer: A, C, D