SATHEE: Complex Numbers 2 Question 16

Complex Numbers 2 Question 16

16.

For positive integers $n_{1}, n_{2}$ the value of expression $(1 + i)^{n_{1}} + {(1 + i^{3})}^{n_{1}} + {(1 + i^{5})}^{n_{2}} + {(1 + i^{7})}^{n_{2}}$ , here $i = \sqrt{- 1}$ is a real number, if and only if

$(1996, 2 M)$

(a) $n_{1} = n_{2} + 1$

(b) $n_{1} = n_{2} - 1$

(d) $n_{1} > 0, n_{2} > 0$

Show Answer

Answer:

Correct Answer: 16. (d)

Solution:

$(1 + i)^{n_{1}} + (1 - i)^{n_{1}} + (1 + i)^{n_{2}} + (1 - i)^{n_{2}}$

$\begin{aligned} = [^{n_{1}} C_{0} +^{n_{1}} C_{1} i +^{n_{1}} C_{2} i^{2} +^{n_{1}} C_{3} i^{3} + \dots] \\ + [^{n_{1}} C_{0} -^{n_{1}} C_{1} i +^{n_{1}} C_{2} i^{2} -^{n_{1}} C_{3} i^{3} + \dots] \\ + [^{n_{2}} C_{0} +^{n_{2}} C_{1} i +^{n_{2}} C_{2} i^{2} +^{n_{2}} C_{3} i^{3} + \dots] \\ + [^{n_{2}} C_{0} -^{n_{2}} C_{1} i +^{n_{2}} C_{2} i^{2} -^{n_{2}} C_{3} i^{3} + . .] \\ = 2 [^{n_{1}} C_{0} +^{n_{1}} C_{2} i^{2} +^{n_{1}} C_{4} i^{4} + \dots] \\ + 2 [^{n_{2}} C_{0} +^{n_{2}} C_{2} i^{2} +^{n_{2}} C_{4} i^{4} + \dots] \\ = 2 [^{n_{1}} C_{0} -^{n_{1}} C_{2} +^{n_{1}} C_{4} - \dots] + 2 [^{n_{2}} C_{0} -^{n_{2}} C_{2} \\ +^{n_{2}} C_{4} - \dots] \end{aligned}$

This is a real number irrespective of the values of $n_{1}$ and $n_{2}$ .

Alternate Solution

$(1 + i)^{n_{1}} + (1 - i)^{n_{1}} + (1 + i)^{n_{2}} + (1 - i)^{n_{2}}$

$\Rightarrow$ A real number for all $n_{1}$ and $n_{2} \in R$ .

$[∵ z + \bar{z} = 2 Re (z) \Rightarrow (1 + i)^{n_{1}} + (1 - i)^{n_{1}}$ is real number for all $n \in R]$

Complex Numbers 2 Question 16

16.

Answer:

Alternate Solution

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Complex Numbers 2 Question 16

16.

Answer:

Alternate Solution

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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