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Complex Numbers question

2014 · Shift 2 · Q27
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Complex Numbers question

2014 · Shift 2 · Q27

JEE AdvancedMathematicsComplex NumbersMCQ+3 / −1
Let zk{z_k}zk​=cos⁡(2kπ10)+i  sin⁡(2kπ10); k=1,2....,9\cos \left( {{{2k\pi } \over {10}}} \right) + i\,\,\sin \left( {{{2k\pi } \over {10}}} \right);\,k = 1,2....,9cos(102kπ​)+isin(102kπ​);k=1,2....,9 List-I P. For each zk{z_k}zk​= there exits as zj{z_j}zj​ such that zk{z_k}zk​. zj{z_j}zj​ = 1 Q. There exists a k∈{1,2,....,9}k \in \left\{ {1,2,....,9} \right\}k∈{1,2,....,9} such that z1.z=zk{z_1}.z = {z_k}z1​.z=zk​ has no solution z in the set of complex numbers R. ∣1−z1∣ ∣1−z2∣ ....∣1−z9∣10{{\left| {1 - {z_1}} \right|\,\left| {1 - {z_2}} \right|\,....\left| {1 - {z_9}} \right|} \over {10}}10∣1−z1​∣∣1−z2​∣....∣1−z9​∣​ equals S. 1−∑k=19cos⁡(2kπ10)1 - \sum\limits_{k = 1}^9 {\cos \left( {{{2k\pi } \over {10}}} \right)}1−k=1∑9​cos(102kπ​) equals List-II 1. True 2. False 3. 1 4. 2
  1. A
    P = 1, Q = 2, R = 4, S = 3
  2. B
    P = 2, Q = 1, R = 3, S = 4
  3. C
    P = 1, Q = 2, R = 3, S = 4
  4. D
    P =2, Q = 1, R = 4, S = 3
View written solutionFree

Correct answer: C

We are given

zk=cos⁡(2kπ10)+isin⁡(2kπ10)=ei2kπ10,k=1,2,…,9z_k=\cos\left(\frac{2k\pi}{10}\right)+i\sin\left(\frac{2k\pi}{10}\right)=e^{i\frac{2k\pi}{10}},\qquad k=1,2,\dots,9zk​=cos(102kπ​)+isin(102kπ​)=ei102kπ​,k=1,2,…,9

These are the non-real/nontrivial 10th roots of unity except z0=1z_0=1z0​=1.

We evaluate each statement in List-I.

1. Statement P

The statement is:

For each zkz_kzk​, there exists a zjz_jzj​ such that zk zj=1.z_k\,z_j=1.zk​zj​=1.

Since zk=ei2kπ/10,z_k=e^{i2k\pi/10},zk​=ei2kπ/10, its multiplicative inverse is zk−1=e−i2kπ/10=ei2(10−k)π/10=z10−k.z_k^{-1}=e^{-i2k\pi/10}=e^{i2(10-k)\pi/10}=z_{10-k}.zk−1​=e−i2kπ/10=ei2(10−k)π/10=z10−k​. Because k∈{1,2,…,9}k\in\{1,2,\dots,9\}k∈{1,2,…,9}, we have 10−k∈{1,2,…,9}10-k\in\{1,2,\dots,9\}10−k∈{1,2,…,9} as well. Hence for every zkz_kzk​, choose zj=z10−k,z_j=z_{10-k},zj​=z10−k​, then zkzj=zkz10−k=ei2kπ/10ei2(10−k)π/10=ei2π=1.z_k z_j=z_k z_{10-k}=e^{i2k\pi/10}e^{i2(10-k)\pi/10}=e^{i2\pi}=1.zk​zj​=zk​z10−k​=ei2kπ/10ei2(10−k)π/10=ei2π=1.

So P is True. Thus, P=1.P=1.P=1.

2. Statement Q

The statement is:

There exists a k∈{1,2,…,9}k\in\{1,2,\dots,9\}k∈{1,2,…,9} such that z1 z=zkz_1\,z=z_kz1​z=zk​ has no solution zzz in the set of complex numbers.

But for any fixed kkk, z=zkz1z=\frac{z_k}{z_1}z=z1​zk​​ is a complex number, since z1≠0z_1\neq 0z1​=0. Therefore the equation always has a complex solution.

Hence the statement "there exists such a kkk with no solution" is False.

So, Q=2.Q=2.Q=2.

3. Statement R

We need to evaluate

∣1−z1∣ ∣1−z2∣⋯∣1−z9∣10.\frac{|1-z_1|\,|1-z_2|\cdots |1-z_9|}{10}.10∣1−z1​∣∣1−z2​∣⋯∣1−z9​∣​.

Use the factorization x10−1=(x−1)∏k=19(x−zk).x^{10}-1=(x-1)\prod_{k=1}^9 (x-z_k).x10−1=(x−1)∏k=19​(x−zk​). Differentiate both sides or use the standard identity for roots of unity:

∏k=19(1−zk)=10.\prod_{k=1}^{9}(1-z_k)=10.k=1∏9​(1−zk​)=10.

Taking modulus on both sides,

∏k=19∣1−zk∣=∣10∣=10.\prod_{k=1}^{9}|1-z_k|=|10|=10.k=1∏9​∣1−zk​∣=∣10∣=10.

Therefore,

∣1−z1∣ ∣1−z2∣⋯∣1−z9∣10=1010=1.\frac{|1-z_1|\,|1-z_2|\cdots |1-z_9|}{10}=\frac{10}{10}=1.10∣1−z1​∣∣1−z2​∣⋯∣1−z9​∣​=1010​=1.

In List-II, 111 corresponds to 3. So, R=3.R=3.R=3.

4. Statement S

We need to evaluate

1−∑k=19cos⁡(2kπ10).1-\sum_{k=1}^9 \cos\left(\frac{2k\pi}{10}\right).1−k=1∑9​cos(102kπ​).

Now,

1+∑k=19zk=01+\sum_{k=1}^9 z_k=01+k=1∑9​zk​=0

because the sum of all 10th roots of unity is 000. Taking real parts,

1+∑k=19cos⁡(2kπ10)=0.1+\sum_{k=1}^9 \cos\left(\frac{2k\pi}{10}\right)=0.1+k=1∑9​cos(102kπ​)=0.

Hence,

∑k=19cos⁡(2kπ10)=−1.\sum_{k=1}^9 \cos\left(\frac{2k\pi}{10}\right)=-1.k=1∑9​cos(102kπ​)=−1.

Therefore,

1−∑k=19cos⁡(2kπ10)=1−(−1)=2.1-\sum_{k=1}^9 \cos\left(\frac{2k\pi}{10}\right)=1-(-1)=2.1−k=1∑9​cos(102kπ​)=1−(−1)=2.

In List-II, 222 corresponds to 4. So, S=4.S=4.S=4.

5. Match with options

We found: P=1,Q=2,R=3,S=4.P=1,\quad Q=2,\quad R=3,\quad S=4.P=1,Q=2,R=3,S=4.

This matches Option C.

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So the stored answer is correct.

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