Complex Numbers
Operations on Complex Numbers
Grade 11
Question:
<p>If <math>z_1 = a + ib</math> and <math>z_2 = c + id</math> are two complex numbers where <math>a, b, c, d \in \mathbb{R}</math> and <math>|z_1| = |z_2| = 1</math> and <math>\text{Im}(z_1 z_2) = 0</math>. If <math>w_1 = a + ic</math> and <math>w_2 = b + id</math>, then:</p>
<p>(a) <math>\text{Im}(w_1 w_2) = 0</math></p>
<p>(b) <math>\text{Im}(w_1 w_2) = 0</math></p>
<p>(c) <math>\text{Im}\left(\frac{w_1}{w_2}\right) = 0</math></p>
<p>(d) <math>\text{Re}\left(\frac{w_1}{w_2}\right) = 0</math></p>
Step-by-Step Solution
Key Concept: Use the constraint that the imaginary part of a product is zero to establish relationships between the coefficients, then apply these to find properties of the new complex numbers.
<p>Given <math>|z_1| = |z_2| = 1</math>, we have <math>a^2 + b^2 = 1</math> and <math>c^2 + d^2 = 1</math>.</p><p><math>z_1 z_2 = (a+ib)(c+id) = (ac-bd) + i(ad+bc)</math></p><p>Since <math>\text{Im}(z_1 z_2) = 0</math>, we have <math>ad + bc = 0</math>, so <math>ad = -bc</math>.</p><p>For <math>w_1 w_2 = (a+ic)(b+id) = (ab-cd) + i(ad+bc) = (ab-cd) + i(0) = ab-cd</math></p><p>Thus <math>\text{Im}(w_1 w_2) = 0</math>. Options (a) and (b) are correct.</p><p><math>\frac{w_1}{w_2} = \frac{a+ic}{b+id} = \frac{(a+ic)(b-id)}{b^2+d^2} = \frac{(ab+cd) + i(cb-ad)}{b^2+d^2}</math></p><p>Since <math>ad = -bc</math>, we have <math>cb - ad = 0</math>.</p><p>Thus <math>\text{Im}\left(\frac{w_1}{w_2}\right) = 0</math>. Option (c) is correct.</p>
Correct Answer: A, B, C