1
JEE Advanced 2023 Paper 1 Online
MCQ (Single Correct Answer)
+3
-1
Let $z$ be a complex number satisfying $|z|^3+2 z^2+4 \bar{z}-8=0$, where $\bar{z}$ denotes the complex conjugate of $z$. Let the imaginary part of $z$ be nonzero.
Match each entry in List-I to the correct entries in List-II.
The correct option is:
Match each entry in List-I to the correct entries in List-II.
List - I | List - II |
---|---|
(P) $|z|^2$ is equal to | (1) 12 |
(Q) $|z-\bar{z}|^2$ is equal to | (2) 4 |
(R) $|z|^2+|z+\bar{z}|^2$ is equal to | (3) 8 |
(S) $|z+1|^2$ is equal to | (4) 10 |
(5) 7 |
The correct option is:
2
JEE Advanced 2021 Paper 1 Online
MCQ (Single Correct Answer)
+3
-1
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,
$$P:\left| {{z_2} - {z_1}} \right| + \left| {{z_3} - {z_2}} \right| + ..... + \left| {{z_{10}} - {z_9}} \right| + \left| {{z_1} - {z_{10}}} \right| \le 2\pi $$
$$Q:\left| {z_2^2 - z_1^2} \right| + \left| {z_3^2 - z_2^2} \right| + .... + \left| {z_{10}^2 - z_9^2} \right| + \left| {z_1^2 - z_{10}^2} \right| \le 4\pi $$
Then,
3
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
4
JEE Advanced 2014 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-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
Questions Asked from Complex Numbers (MCQ (Single Correct Answer))
Number in Brackets after Paper Indicates No. of Questions
JEE Advanced 2023 Paper 1 Online (1)
JEE Advanced 2021 Paper 1 Online (1)
JEE Advanced 2019 Paper 1 Offline (1)
JEE Advanced 2014 Paper 2 Offline (1)
JEE Advanced 2013 Paper 2 Offline (2)
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IIT-JEE 1983 (2)
IIT-JEE 1982 (2)
IIT-JEE 1981 (1)
IIT-JEE 1980 (1)
IIT-JEE 1979 (1)
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