Complex Numbers
Algebraic Operations on Complex Numbers
Grade 11
Question:
<p>If the complex number \(z\) satisfy the equation \((i-z)(1+2i)+(1-iz)(3-4i)=1+7i\), then</p>
<p>There are two values of \(z\) exist</p>
<p>At least one value of \(z\) lies in 1st quadrant</p>
<p>Exactly one value of \(z\) lies in 3rd quadrant</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Expand both products systematically, separate real and imaginary parts, then solve the resulting system of linear equations in terms of the real and imaginary components of z.
<p><strong>Step 1:</strong> Let z = x + iy where x, y are real. Substitute into the equation.</p><p><strong>Step 2:</strong> Expand (i-z)(1+2i) = i + 2i² - x - 2xi = i - 2 - x - 2xi = (-2-x) + (1-2x)i</p><p><strong>Step 3:</strong> Expand (1-iz)(3-4i) = 3 - 4i - 3iz + 4i²z = 3 - 4i - 3iz - 4z</p><p>Substitute z = x + iy: = 3 - 4i - 3i(x+iy) - 4(x+iy) = 3 - 4i - 3ix + 3y - 4x - 4iy = (3+3y-4x) + (-4-3x-4y)i</p><p><strong>Step 4:</strong> Add the two products: [(-2-x) + (3+3y-4x)] + [(1-2x) + (-4-3x-4y)]i = (1+2y-5x) + (-3-5x-4y)i</p><p><strong>Step 5:</strong> Set equal to 1 + 7i:</p><p>Real part: 1 + 2y - 5x = 1 → 2y - 5x = 0 → y = 5x/2</p><p>Imaginary part: -3 - 5x - 4y = 7 → -5x - 4y = 10</p><p><strong>Step 6:</strong> Substitute y = 5x/2 into -5x - 4y = 10:</p><p>-5x - 4(5x/2) = 10 → -5x - 10x = 10 → -15x = 10 → x = -2/3</p><p>Then y = 5(-2/3)/2 = -5/3</p><p>∴ z = -2/3 - 5i/3 or z = (-2-5i)/3</p>
Correct Answer: B