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TS EAMCET 19-July-2022 Shift 1 Solved Paper

Section: Mathematics

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Question : 7 of 160

Marks: +1, -0

If

(2 x-y+1)+i(x-2 y-1)=2-3 i

, then the multiplicative inverse of

(x-i y)

is

$\;\frac{15}{41}+\;\frac{12}{41} i$
$\;\frac{6}{29}+\;\frac{15}{29} i$
$\;\frac{15}{29}+\;\frac{6}{29} i$
$\;\frac{12}{41}+\;\frac{15}{41} i$

Solution: 👈: Video Solution

Given,

(2 x-y+1)+i(x-2 y-1)=2-3 i

\Rightarrow 2 x-y+1=2

and

x-2 y-1=-3

\Rightarrow 2 x-y=1

and

x=2 y-2

Now, substitute value of

x=2 y-2

in equation

2 x-y=1

, we get

2(2 y-2)-y=1 \Rightarrow 4 y-4-y=1

3 y=5 \Rightarrow y=\;\frac{5}{3}

Now, substitute

y=\;\frac{5}{3}

in

x=2 y-2

x=\;\frac{10}{3}-2=\;\frac{4}{3}

Multiplicative inverse of

(x-i y)=\;\frac{1}{x-i y}

\;=\;\frac{1}{\frac{4}{3}-i\;\frac{5}{3}}=\;\frac{3}{4-5 i} \times \;\frac{(4+5 i)}{(4+5 i)}

\;=\;\frac{12}{4^2+5^2}+\;\frac{15}{4^2+5^2} i=\;\frac{12}{41}+\;\frac{15}{41} i

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