NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### JEE Main 2022 (Online) 27th June Morning Shift [ Memory Based ]

If the non-real roots of z2 = iz the vertices of a polygon then the area of the polygon is

A
$${{\sqrt 3 } \over 4}$$
B
$${{\sqrt 3 } \over 2}$$
C
$${{\sqrt 5 } \over 2}$$
D
$${{\sqrt 5 } \over 4}$$
2

### JEE Main 2022 (Online) 25th June Evening Shift [ Memory Based ]

If z1 and z2 are two complex number such that $${\overline z _1} = i{\overline z _2}$$ and $$\arg \left( {{{{z_1}} \over {{{\overline z }_2}}}} \right) = \pi$$ then

A
$$\arg {z_1} = {{3\pi } \over 4}$$
B
$$\arg {z_2} = {{ - 3\pi } \over 4}$$
C
$$\arg {z_1} = {\pi \over 4}$$
D
$$\arg {z_2} = {{ - \pi } \over 4}$$
3

### JEE Main 2021 (Online) 31st August Evening Shift

If z is a complex number such that $${{z - i} \over {z - 1}}$$ is purely imaginary, then the minimum value of | z $$-$$ (3 + 3i) | is :
A
$$2\sqrt 2 - 1$$
B
$$3\sqrt 2$$
C
$$6\sqrt 2$$
D
$$2\sqrt 2$$

## Explanation

$${{z - i} \over {z - 1}}$$ is purely imaginary number

Let $$z = x + iy$$

$$\therefore$$ $${{x + i(y - 1)} \over {(x - 1) + i(y)}} \times {{(x - 1) - iy} \over {(x - 1) - iy}}$$

$$\Rightarrow {{x(x - 1) + y(y - 1) + i( - y - x + 1)} \over {{{(x - 1)}^2} + {y^2}}}$$ is purely imaginary number

$$\Rightarrow x(x - 1) + y(y - 1) = 0$$

$$\Rightarrow {\left( {x - {1 \over 2}} \right)^2} + {\left( {y - {1 \over 2}} \right)^2} = {1 \over 2}$$

$$\therefore$$ $${\left| {z - (3 + 3i)} \right|_{\min }} = \left| {PC} \right| - {1 \over {\sqrt 2 }}$$

$$= {5 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }} = 2\sqrt 2$$
4

### JEE Main 2021 (Online) 27th August Morning Shift

If $$S = \left\{ {z \in C:{{z - i} \over {z + 2i}} \in R} \right\}$$, then :
A
S contains exactly two elements
B
S contains only one element
C
S is a circle in the complex plane
D
S is a straight line in the complex plane

## Explanation

Given $${{z - i} \over {z + 2i}} \in R$$

Then $$\arg \left( {{{z - i} \over {z + 2i}}} \right)$$ is 0 or $$\pi$$

$$\Rightarrow$$ S is straight line in complex

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12