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UPSC NDA Math Model Paper 8

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Question : 103 of 120

Marks: +1, -0

What is the square root of (8 - 6i), Where

\sqrt{-1} = i

?

± (2 - i)
± (3 + i)
± (3 - i)
None of the above

Solution:

Let

\sqrt{8-6i} = x + iy

Squaring both sides, we get

\Rightarrow 8-6i = (x+iy)^2

\Rightarrow 8-6i = x^{2}+(iy)^{2}+i 2 x y

\Rightarrow 8-6i = x^{2}-y^{2}+i 2 x y

Compare real and imaginary parts,

\therefore x^{2}-y^{2}=8

and

2 x y = -6

2 xy = -6 \Rightarrow y = \frac{-3}{x}

.......(1)

Now,

x^{2}-y^{2}=8

\Rightarrow x^{2}-\frac{9}{x^{2}}=8

\Rightarrow x^{4}-9=8x^{2}

\Rightarrow x^{4}-8x^{2}-9=0

\Rightarrow x^{4}-9x^{2}+x^{2}-9=0

\Rightarrow x^{2}(x^{2}-9)+1(x^{2}-9)=0

\Rightarrow (x^{2}+1)(x^{2}-9)=0

So,

(x^{2}+1)=0 \Rightarrow x^{2} \neq -1

And

(x^{2}-9)=0

\therefore x = \pm 3

Put the value of

x

in equation

1^{\text{st}}

If

x=3

than

y=-1

If

x=-3

than

y=1

So, square root of

(8-6i) = (3-i)

or

(-3+i)

\therefore \sqrt{8-6i} = \pm (3-i)

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