Complex Numbers
Modulus and Arguments
Grade 11
Question:
<p>Let <em>Z</em><sub>1</sub> and <em>Z</em><sub>2</sub> are two non-zero complex number such that <span style="text-decoration:overline;">|<em>Z</em><sub>1</sub> + <em>Z</em><sub>2</sub>| = |<em>Z</em><sub>1</sub>| = |<em>Z</em><sub>2</sub>|</span>, then <em>Z</em><sub>1</sub>/<em>Z</em><sub>2</sub> may be:</p>
<p>(a) \(1 + \omega\)</p>
<p>(b) \(1 + \omega^2\)</p>
<p>(c) \(\omega\)</p>
<p>(d) \(\omega^2\)</p>
Step-by-Step Solution
Key Concept: Use the modulus condition to establish that $Z_1/Z_2$ satisfies a specific polynomial equation related to cube roots of unity.
<p><strong>Key insight:</strong> Use the condition $|Z_1 + Z_2| = |Z_1| = |Z_2|$ to find the relationship between $Z_1$ and $Z_2$.</p><p>Let $|Z_1| = |Z_2| = r$. Then $|Z_1 + Z_2| = r$.</p><p>$|Z_1 + Z_2|^2 = r^2$ implies $(Z_1 + Z_2)(\overline{Z_1} + \overline{Z_2}) = r^2$</p><p>$|Z_1|^2 + |Z_2|^2 + Z_1\overline{Z_2} + Z_2\overline{Z_1} = r^2$</p><p>$r^2 + r^2 + 2\text{Re}(Z_1\overline{Z_2}) = r^2$</p><p>$\text{Re}(Z_1\overline{Z_2}) = -\frac{r^2}{2}$</p><p>This means $\frac{Z_1}{Z_2}$ has real part $-\frac{1}{2}$ when normalized. The solutions are the cube roots of unity shifted: $1 + \omega$, $1 + \omega^2$, $\omega^2$.</p>
Correct Answer: a,b,d