Consider the following in respect of a complex number Z : 1. (Z^ -1 ) =( Z )^ -1 2. Z Z^ -1 =|Z|^2 Which of the above is/are correct?

Correct answer: C. Both 1 and 2 Let z=x+i y, z =x-i y and z^ -1 =(1)/(z)=(1)/(x+i y) 1. ( z^ -1 )=( (1)/(x+i y) )=( (x-i y)/(x^2+y^2) )=((x+i y)/(x^2+y^2)) aligned & ( z )^ -1 =(x-i y)^ -1 =(1)/(x-i y)=(x+i y)/(x^2+y^2) & ( z^ -1 )=( z )^ -1 aligned 2. z · z^ -1 =(x+i y)((1)/(x-i y))=((x+i y)(x+i y))/((x-i y)(x+i y))=((x+i y)^2)/(x^2+y^2)=|z|^2 Hence, both (1) and (2) are true.

NDA 2021

Exam: NDA
Year: 2021
Subject: Mathematics
Chapter: Complex Number

NDA 2021 · Mathematics · Complex Number

Consider the following in respect of a complex number Z : 1. (Z^ -1...

Consider the following in respect of a complex number

Z

: 1.

\overline{\left(Z^{-1}\right)}=(\bar{Z})^{-1} 2.

Z Z^{-1}=|Z|^2

Which of the above is/are correct?

Options

A $1$ only
B $2$ only
CBoth $1$ and $2$
DNeither $1$ nor $2$

Correct answer

C. Both $1$ and $2$

Step-by-step solution

Let

z=x+i y, \bar{z}=x-i y and

z^{-1}=\frac{1}{z}=\frac{1}{x+i y} 1. \left(\overline{z^{-1}}\right)=\left(\overline{\frac{1}{x+i y}}\right)=\left(\overline{\frac{x-i y}{x^2+y^2}}\right)=\left(\frac{x+i y}{x^2+y^2}\right)

\begin{aligned} & (\bar{z})^{-1}=(x-i y)^{-1}=\frac{1}{x-i y}=\frac{x+i y}{x^2+y^2} \\ & \therefore\left(\overline{z^{-1}}\right)=(\bar{z})^{-1}\end{aligned}

2. \quad z \cdot z^{-1}=(x+i y)\left(\frac{1}{x-i y}\right)=\frac{(x+i y)(x+i y)}{(x-i y)(x+i y)}=\frac{(x+i y)^2}{x^2+y^2}=|z|^2

Hence, both (1) and (2) are true.

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