1
JEE Advanced 2021 Paper 1 Online
MCQ (Single Correct Answer)
+3
-1
Let $\theta_1, \theta_2, \ldots, \theta_{10}$ be positive valued angles (in radian) such that $\theta_1+\theta_2+\cdots+\theta_{10}=2 \pi$. Define the complex numbers $z_1=e^{i \theta_1}, z_k=z_{k-1} e^{i \theta_k}$ for $k=2,3, \ldots, 10$, 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,
$$P:\left| {{z_2} - {z_1}} \right| + \left| {{z_3} - {z_2}} \right| + ..... + \left| {{z_{10}} - {z_9}} \right| + \left| {{z_1} - {z_{10}}} \right| \le 2\pi $$
$$Q:\left| {z_2^2 - z_1^2} \right| + \left| {z_3^2 - z_2^2} \right| + .... + \left| {z_{10}^2 - z_9^2} \right| + \left| {z_1^2 - z_{10}^2} \right| \le 4\pi $$
Then,
2
JEE Advanced 2019 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1
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
3
JEE Advanced 2014 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
Let $${z_k}$$ = $$\cos \left( {{{2k\pi } \over {10}}} \right) + i\,\,\sin \left( {{{2k\pi } \over {10}}} \right);\,k = 1,2....,9$$
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
1. True
2. False
3. 1
4. 2
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
4
JEE Advanced 2013 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-1
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 $${S_3} = \left\{ {z \in C:{\mathop{\rm Re}\nolimits} z > 0} \right\}\,$$.
$$\,\mathop {\min }\limits_{z \in S} \left| {1 - 3i - z} \right| = $$
Questions Asked from Complex Numbers (MCQ (Single Correct Answer))
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