Question

The point of intersection of the curves arg (z – 3i) = 3p/4 and arg(2z + 1 – 2i) = p/4 is –

• A. $\frac{1}{4}$ (3 + 9i)
• B. $\frac{1}{4}$ (9 + 3i)
• C. 1 + I
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

Sol. arg (z – 3i) = arg (x + (y – 3) i) = 3p/4

Ž x < 0, y – 3 > 0, $\frac{y - 3}{x}$ = tan $\frac{3\pi}{4}$= – 1

Ž x + y – 3 = 0, x < 0, y > 3. …(1)

arg (2z + 1 – 2i) = arg {(2x + 1) + i(2y – 2)} = p/4

Ž x > –1/2, y > 1 and $\frac{2(y–1)}{2x + 1}$ = tan$\frac{\pi}{4}$= 1

Ž 2x – 2y + 3 = 0, x > –1/2, y > 1 …(2)

Hence, (1) and (2) have no point of intersection.