Complex Numbers
Complex Arithmetic
Grade 11
Question:
<p>If <span>\(Z = \frac{7 + i}{3 + 4i}\)</span>, then find <span>\(Z^{14}\)</span>:</p>
<p>(a) <span>\(2^7\)</span></p>
<p>(b) <span>\((-2)^7\)</span></p>
<p>(c) <span>\((2^7)i\)</span></p>
<p>(d) <span>\((-2^7)i\)</span></p>
Step-by-Step Solution
Key Concept: Simplify the complex number Z to a form whose powers follow a pattern, then use De Moivre's theorem or direct exponentiation. Notice that Z simplifies to a purely imaginary number with magnitude 1/5, which when raised to the 14th power yields a real multiple of i.
<p><strong>Step 1: Simplify Z by multiplying by the conjugate.</strong></p><p>Multiply numerator and denominator by the conjugate of (3 + 4i), which is (3 - 4i):</p><p>$$Z = \frac{7 + i}{3 + 4i} \cdot \frac{3 - 4i}{3 - 4i}$$</p><p><strong>Step 2: Calculate the numerator.</strong></p><p>$$(7 + i)(3 - 4i) = 21 - 28i + 3i - 4i^2 = 21 - 25i + 4 = 25 - 25i$$</p><p><strong>Step 3: Calculate the denominator.</strong></p><p>$$(3 + 4i)(3 - 4i) = 9 - 16i^2 = 9 + 16 = 25$$</p><p><strong>Step 4: Simplify Z.</strong></p><p>$$Z = \frac{25 - 25i}{25} = 1 - i$$</p><p><strong>Step 5: Convert to polar form or use exponentiation directly.</strong></p><p>$$|Z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$$</p><p>$$\arg(Z) = -\frac{\pi}{4}$$ (since Z is in the fourth quadrant)</p><p>$$Z = \sqrt{2} e^{-i\pi/4}$$</p><p><strong>Step 6: Compute Z¹⁴.</strong></p><p>$$Z^{14} = (\sqrt{2})^{14} e^{-i\pi \cdot 14/4} = 2^7 e^{-i7\pi/2}$$</p><p><strong>Step 7: Simplify the exponential.</strong></p><p>$$e^{-i7\pi/2} = \cos(-7\pi/2) + i\sin(-7\pi/2)$$</p><p>Since $-7\pi/2 = -2\pi - 3\pi/2$, we have:</p><p>$$\cos(-7\pi/2) = \cos(-3\pi/2) = 0$$</p><p>$$\sin(-7\pi/2) = \sin(-3\pi/2) = 1$$</p><p>Therefore: $e^{-i7\pi/2} = i$</p><p><strong>Step 8: Final result.</strong></p><p>$$Z^{14} = 2^7 \cdot i = 128i$$</p><p>∴ Answer: C</p>
Correct Answer: C