Find the square roots of 9-12i 

  • Option 1)

    \pm (2\sqrt3 - \sqrt 3 \:i)

  • Option 2)

    \pm (3 + 2i)

  • Option 3)

    \pm (\sqrt3 - 2\sqrt 3 \:i)

  • Option 4)

    \pm (3 - 2i)

 

Answers (1)

As learnt in concept

Definition of Complex Number -

z=x+iy, x,y\epsilon R  & i2=-1

- wherein

Real part of z = Re (z) = x & Imaginary part of z = Im (z) = y

 

 \sqrt{9-12i}= x+iy

9-12i= (x^{2}-y^{2})+2ixy

(x^{2}-y^{2})=9\: \: and \: \: xy=-6

here, x= 2\sqrt3\: \: and \: \: y= -\sqrt3

Thus square root = \pm (2\sqrt3- \sqrt 3i)

 


Option 1)

\pm (2\sqrt3 - \sqrt 3 \:i)

Correct option

Option 2)

\pm (3 + 2i)

Incorrect option 

Option 3)

\pm (\sqrt3 - 2\sqrt 3 \:i)

Incorrect option 

Option 4)

\pm (3 - 2i)

Incorrect option 

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