Point of intersection of the lines arg (z-1)=(π/4) and z=i t+(1-t) is z=i t+(1-t) is

Question

Point of intersection of the lines arg (z-1)=(π/4) and z=i t+(1-t) is z=i t+(1-t) is (A) i (B) 1 (C) 1+i (D) None of these

Tardigrade Admin · Accepted Answer

arg (z-1)=(π/4) i.e., a ray with initial point (1,0) however excluding this point and inclined at an angle 45° with x-axis. z=i t+(1-t) i.e., x=1-t and y=t or x+y=1 Clearly, point of intersection is (1,0). But (1,0) does not satisfy the first equation. Hence, no point of intersection.

Q. Point of intersection of the lines arg $(z - 1) = \frac{π}{4}$ and $z = i t + (1 - t)$ is $z = i t + (1 - t)$ is

$i$

$1$

$1 + i$

None of these

Q. Point of intersection of the lines arg (z−1)=4π​ and z=it+(1−t) is z=it+(1−t) is

i

1

1+i

None of these

arg(z−1)=4π​ i.e., a ray with initial point (1,0) however excluding this point and inclined at an angle 45∘ with x-axis. z=it+(1−t) i.e., x=1−t and y=t or x+y=1 Clearly, point of intersection is (1,0). But (1,0) does not satisfy the first equation. Hence, no point of intersection.

Q. Point of intersection of the lines arg $(z - 1) = \frac{π}{4}$ and $z = i t + (1 - t)$ is $z = i t + (1 - t)$ is

$i$

$1$

$1 + i$