Complex Numbers Ques 20

For positive integers $n_1, n_2$ the value of expression $(1+i)^{n_1}+\left(1+i^{3}\right)^{n_1}+\left(1+i^{5}\right)^{n_2}+\left(1+i^{7}\right)^{n_2}$, here $i=\sqrt{-1}$ is a real number, if and only if

$(1996,2 M)$

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

(b) $n_1=n_2-1$

(c) $n _1=n _2$

(d) $n _1>0, n _2>0$

Show Answer

Answer:

Correct Answer: 20.(d)

Solution:

Formula:

The complex number system:

  1. $(1+i)^{n _1}+(1-i)^{n _1}+(1+i)^{n _2}+(1-i)^{n _2}$

$ \begin{aligned} & =\left[{ }^{n _1} C _0+{ }^{n _1} C _1 i+{ }^{n _1} C _2 i^{2}+{ }^{n _1} C _3 i^{3}+\ldots\right] \\ & +\left[{ }^{n _1} C _0-{ }^{n _1} C _1 i+{ }^{n _1} C _2 i^{2}-{ }^{n _1} C _3 i^{3}+\ldots\right] \\ & +\left[{ }^{n _2} C _0+{ }^{n _2} C _1 i+{ }^{n _2} C _2 i^{2}+{ }^{n _2} C _3 i^{3}+\ldots\right] \\ & +\left[{ }^{n _2} C _0-{ }^{n _2} C _1 i+{ }^{n _2} C _2 i^{2}-{ }^{n _2} C _3 i^{3}+. .\right] \\ & =2\left[{ }^{n _1} C _0+{ }^{n _1} C _2 i^{2}+{ }^{n _1} C _4 i^{4}+\ldots\right] \\ & +2\left[{ }^{n _2} C _0+{ }^{n _2} C _2 i^{2}+{ }^{n _2} C _4 i^{4}+\ldots\right] \\ & =2\left[{ }^{n_1} C_0-{ }^{n_1} C_2+{ }^{n_1} C_4-\ldots\right]+2\left[{ }^{n_2} C_0-{ }^{n_2} C_2\right. \\ & \left.+{ }^{n_2} C_4-\ldots\right] \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 $n_2 \in \mathbb{R}$.

$\left[\because z+\bar{z}=2 \operatorname{Re}(z) \Rightarrow(1+i)^{n}+(1-i)^{n}\right.$ is real number for all $n \in \mathbb{N}]$



Table of Contents

sathee Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on

Please select your preferred language
कृपया अपनी पसंदीदा भाषा चुनें