Complex Numbers
Argument of complex number
Grade 11
Question:
<p>If \(x = 9^{1/3} \times 9^{1/9} \times 9^{1/27} \times \cdots\); \(y = 4^{1/3} \times 4^{-1/9} \times 4^{1/27} \times \cdots\); and \(z = \sum_{r=1}^{\infty}(1+i)^{-r}\), then \(\arg(x + yz)\) is equal to</p>
<p>(1) 0</p>
<p>(2) \(\pi - \tan^{-1}\!\left(\dfrac{\sqrt{2}}{3}\right)\)</p>
<p>(3) \(-\tan^{-1}\!\left(\dfrac{\sqrt{2}}{3}\right)\)</p>
<p>(4) \(-\tan^{-1}\!\left(\dfrac{2}{\sqrt{3}}\right)\)</p>
Step-by-Step Solution
Key Concept: Recognize that x and y are infinite products with exponents forming geometric series; evaluate each using the formula ∑(1/3^(n-1)) and ∑((-1)^(n-1)/3^(n-1)). For z, sum the geometric series ∑(1+i)^(-r) directly, then find the argument of the final complex number.
<p><strong>Step 1: Find x</strong></p><p>x = 9^(1/3 + 1/9 + 1/27 + ...)</p><p>Exponent is geometric series: S₁ = (1/3)/(1 - 1/3) = 1/2</p><p>Therefore, x = 9^(1/2) = 3</p><p><strong>Step 2: Find y</strong></p><p>y = 4^(1/3 - 1/9 + 1/27 - ...)</p><p>Exponent is geometric series: S₂ = (1/3)/(1 + 1/3) = 1/4</p><p>Therefore, y = 4^(1/4) = √2</p><p><strong>Step 3: Find z</strong></p><p>z = ∑(1+i)^(-r) = ∑[1/(1+i)]^r</p><p>This is geometric series with first term a = 1/(1+i) and ratio r = 1/(1+i)</p><p>z = [1/(1+i)]/(1 - 1/(1+i)) = [1/(1+i)]/[i/(1+i)] = 1/i = -i</p><p><strong>Step 4: Calculate x + yz</strong></p><p>x + yz = 3 + √2·(-i) = 3 - i√2</p><p><strong>Step 5: Find arg(3 - i√2)</strong></p><p>Since 3 - i√2 is in fourth quadrant (positive real, negative imaginary)</p><p>arg(3 - i√2) = arctan(-√2/3) = -arctan(√2/3)</p><p>∴ Answer: <strong>-arctan(√2/3)</strong> or equivalent form matching option B</p>
Correct Answer: B