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  • Solve it, - Complex numbers and quadratic equations - JEE Main

The equation \small Im\left ( \frac{iz -2}{z - i} \right ) + 1 = 0, z\: \epsilon \; C, z \neq i represents a part of a circle having radius equal to :
 

  • Option 1)

    2

  • Option 2)

    1

  • Option 3)

    \frac{3}{4}

  • Option 4)

    \frac{1}{2}

 

Answers (2)

Using the definition of complex numbers as we studied in Definition of Complex Number -

z=x+iy, x,y\epsilon R  & i2=-1

-

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

 we can say, \frac{iz-2}{z-i}=\frac{ix-y-2}{x+iy-i}=\frac{-y-2+ix}{x+i(y-1)}

Proceeding further by Division of Complex Numbers -

\frac{a+ib}{c+id}=\frac{ac+bd}{c^{2}+d^{2}}+i\frac{bc-ad}{c^{2}+d^{2}}

-  

We have, \frac{-yx-2x+xy-x}{x^{2}+(y-1)^{2}}+i\frac{x^{2}+y^{2}+y-2}{x^{2}+(y-1)^{2}}

It has been given that the Imaginary part of this complex number is -1.

Therefore, \frac{x^{2}+y^{2}+y-2}{x^{2}+(y-1)^{2}}=-1\Rightarrow 2x^{2}+2y^{2}-y-1=0\Rightarrow x^{2}+y^{2}-(1/2)y-(1/2)=0

Whose radius is:\sqrt{\frac{1}{16}+\frac{1}{2}}=\sqrt{\frac{9}{16}}=\frac{3}{4}

 


Option 1)

2

This is a wrong option.

Option 2)

1

This is a wrong option.

Option 3)

\frac{3}{4}

This is the correct option.

Option 4)

\frac{1}{2}

This is a wrong option.

Posted by

Sabhrant Ambastha

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