Complex Numbers
Polynomial division and remainder theorem
Grade 11
Question:
<p>Dividing \(f(z)\) by \(z - i\), we obtain the remainder \(i\) and dividing it by \(z + i\), we get the remainder \(1 + i\), then the remainder upon the division of \(f(z)\) by \(z^2 + 1\) is</p>
<p>\(\dfrac{1}{2}(z+1)+i\)</p>
<p>\(\dfrac{1}{2}(iz+1)+i\)</p>
<p>\(\dfrac{1}{2}(iz-1)+i\)</p>
<p>\(\dfrac{1}{2}(z+i)+1\)</p>
Step-by-Step Solution
Key Concept: Use the Remainder Theorem: f(i) = i and f(-i) = 1+i. Since we're dividing by z²+1 = (z-i)(z+i), the remainder must be a linear polynomial az+b. Substitute the two conditions to find a and b.
<p><strong>Step 1:</strong> By Remainder Theorem, f(i) = i and f(-i) = 1+i</p><p><strong>Step 2:</strong> Let the remainder be R(z) = az + b when f(z) is divided by z²+1 = (z-i)(z+i). Then f(z) = Q(z)(z²+1) + az + b</p><p><strong>Step 3:</strong> Applying the remainder theorem conditions:</p><p>At z = i: f(i) = a(i) + b = ai + b = i</p><p>At z = -i: f(-i) = a(-i) + b = -ai + b = 1+i</p><p><strong>Step 4:</strong> From the two equations:</p><p>ai + b = i ... (1)</p><p>-ai + b = 1+i ... (2)</p><p><strong>Step 5:</strong> Adding (1) and (2): 2b = i + 1 + i = 1 + 2i, so b = ½ + i</p><p><strong>Step 6:</strong> Subtracting (2) from (1): 2ai = i - (1+i) = -1, so ai = -½, thus a = i/(-½) = -2i</p><p><strong>Step 7:</strong> Therefore, R(z) = -2iz + (½ + i) = -2iz + ½ + i</p><p>∴ Answer: A</p>
Correct Answer: A