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1

### JEE Advanced 2021 Paper 1 Online

Let $$\theta$$1, $$\theta$$2, ........, $$\theta$$10 = 2$$\pi$$. Define the complex numbers z1 = ei$$\theta$$1, zk = zk $$-$$ 1ei$$\theta$$k for k = 2, 3, ......., 10, where i = $$\sqrt { - 1}$$. Consider the statements P and Q given below :

$$P:\left| {{z_2} - {z_1}} \right| + \left| {{z_3} - {z_2}} \right| + ..... + \left| {{z_{10}} - {z_9}} \right| + \left| {{z_1} - {z_{10}}} \right| \le 2\pi$$

$$Q:\left| {z_2^2 - z_1^2} \right| + \left| {z_3^2 - z_2^2} \right| + .... + \left| {z_{10}^2 - z_9^2} \right| + \left| {z_1^2 - z_{10}^2} \right| \le 4\pi$$

Then,
A
P is TRUE and Q is FALSE
B
Q is TRUE and P is FALSE
C
both P and Q are TRUE
D
both P and Q are FALSE

## Explanation

Both P and Q are true.

$$\because$$ Length of direct ditance $$\le$$ length of arc

i.e. | z2 $$-$$ z1 | = length of line AB $$\le$$ length of arc AB.

| z3 $$-$$ z2 | = length of line BC $$\le$$ length of arc BC.

$$\therefore$$ Sum of length of these 10 lines $$\le$$ sum of length of arcs (i.e. 2$$\pi$$) (because $$\theta$$1 + $$\theta$$2 + $$\theta$$3 + .... + $$\theta$$10 = 2$$\pi$$ given)

$$\therefore$$ | z2 $$-$$ z1 | + | z3 $$-$$ z2 | + ..... + | z1 $$-$$ z10 | $$\le$$ 2$$\pi$$ $$\to$$ P is true.

And | z$$_k^2$$ $$-$$ z$$_{k - 1}^2$$ | = | zk $$-$$ zk $$-$$ 1 | | zk + zk $$-$$ 1 |

As we know that,

| zk + zk $$-$$ 1 | $$\le$$ | zk | + | zk $$-$$ 1 | $$\le$$ 2

$$\therefore$$ | z$$_2^2$$ $$-$$ z$$_1^2$$ | + | z$$_3^2$$ $$-$$ z$$_2^2$$ | + .... + | z$$_1^2$$ $$-$$ z$$_{10}^2$$ | $$\le$$ 2 ( | z2 $$-$$ z1 | + | z3 $$-$$ z2 | + .... + | z1 $$-$$ z10 | )

$$\le$$ 2(2$$\pi$$)

$$\le$$ 4$$\pi$$ $$\to$$ Q is true.
2

### JEE Advanced 2019 Paper 1 Offline

Let S be the set of all complex numbers z satisfying $$\left| {z - 2 + i} \right| \ge \sqrt 5$$. If the complex number z0 is such that $${1 \over {\left| {{z_0} - 1} \right|}}$$ is the maximum of the set $$\left\{ {{1 \over {\left| {{z_0} - 1} \right|}}:z \in S} \right\}$$, then the principal argument of $${{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}}$$ is
A
$${\pi \over 4}$$
B
$${3\pi \over 4}$$
C
$$-$$$${\pi \over 2}$$
D
$${\pi \over 2}$$

## Explanation

The complex number z satisfying $$\left| {z - 2 + i} \right| \ge \sqrt 5$$, which represents the region outside the circle (including the circumference) having centre (2, $${ - 1}$$) and radius $$\sqrt 5$$ units.

Now, for $${{z_0} \in S{1 \over {\left| {{z_0} - 1} \right|}}}$$ is maximum.

When $${\left| {{z_0} - 1} \right|}$$ is minimum. And for this it is required that $${{z_0} \in S}$$, such that z0 is collinear with the points (2, $$-$$1) and (1, 0) and lies on the circumference of the circle $$\left| {z - 2 + i} \right|$$ = $$\sqrt 5$$.

So let z0 = x + iy, and from the figure 0 < x < 1 and y >0.

So, $${{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}} = {{4 - x - iy - x + iy} \over {x + iy - x + iy + 2i}} = {{2(2 - x)} \over {2i(y + 1)}} = - i\left( {{{2 - x} \over {y + 1}}} \right)$$

$$\because$$ $${{{2 - x} \over {y + 1}}}$$ is a positive real number, so $${{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}}$$ is purely negative imaginary number.

$$\Rightarrow$$ $$\arg \left( {{{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}}} \right) = - {\pi \over 2}$$
3

### JEE Advanced 2014 Paper 2 Offline

Let $${z_k}$$ = $$\cos \left( {{{2k\pi } \over {10}}} \right) + i\,\,\sin \left( {{{2k\pi } \over {10}}} \right);\,k = 1,2....,9$$

List-I

P. For each $${z_k}$$ = there exits as $${z_j}$$ such that $${z_k}$$.$${z_j}$$ = 1
Q. There exists a $$k \in \left\{ {1,2,....,9} \right\}$$ such that $${z_1}.z = {z_k}$$ has no solution z in the set of complex numbers
R. $${{\left| {1 - {z_1}} \right|\,\left| {1 - {z_2}} \right|\,....\left| {1 - {z_9}} \right|} \over {10}}$$ equals
S. $$1 - \sum\limits_{k = 1}^9 {\cos \left( {{{2k\pi } \over {10}}} \right)}$$ equals

List-II

1. True
2. False
3. 1
4. 2
A
P = 1, Q = 2, R = 4, S = 3
B
P = 2, Q = 1, R = 3, S = 4
C
P = 1, Q = 2, R = 3, S = 4
D
P =2, Q = 1, R = 4, S = 3
4

### JEE Advanced 2013 Paper 2 Offline

Let $$S = {S_1} \cap {S_2} \cap {S_3}$$, where $${S_1} = \left\{ {z \in C:\left| z \right| < 4} \right\},{S_2} = \left\{ {z \in C:{\mathop{\rm Im}\nolimits} \left[ {{{z - 1 + \sqrt 3 i} \over {1 - \sqrt 3 i}}} \right] > 0} \right\}$$ and $${S_3} = \left\{ {z \in C:{\mathop{\rm Re}\nolimits} z > 0} \right\}\,$$.

$$\,\mathop {\min }\limits_{z \in S} \left| {1 - 3i - z} \right| =$$

A
$${{2 - \sqrt 3 } \over 2}$$
B
$${{2 + \sqrt 3 } \over 2}$$
C
$${{3 - \sqrt 3 } \over 2}$$
D
$${{3 + \sqrt 3 } \over 2}$$

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