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If Re \left [ \frac{z-1}{2z+i} \right ]=1, where z=x+iy, then the point (x,y) lies on a :

Option: 1 circle whose centre is at \left [ -\frac{1}{2},-\frac{3}{2} \right ].

Option: 2 straight line whose slope is \frac{3}{2}.

Option: 3 circle whose diameter is \frac{\sqrt{5}}{2}.

Option: 4 straight line whose slope is -\frac{2}{3}.

Answers (1)




Conjugate of complex numbers and their properties -

The complex conjugate of a complex number a + ib (a, b are real numbers and b ≠ 0) is a − ib. 

It is denoted as  \bar{z}.

i.e. if z = a + ib, then its conjugate is  \bar{z}  = a - ib.

Conjugate of complex numbers is obtained by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.


  • When a complex number is added to its complex conjugate, the result is a real number. i.e. z = a + ib, \bar{z}  = a - ib

Then the sum, z + \bar{z}= a + ib + a - ib = 2a (which is real)

  • When a complex number is multiplied by its complex conjugate, the result is a real number i.e. z = a + ib, \bar{z} = a - ib

Then the product, z?\bar{z} = (a + ib)?(a - ib) = a2 - (ib)2

                                                  = a2 +  b2 (which is real)


Circle(Definition) -

General Form:

The equation of a circle with centre at (h,k) and radius r is 

\\ {\Rightarrow(x-h)^{2}+(y-k)^{2}=r^{2}} \\ {\Rightarrow x^{2}+y^{2}-2 h x-2 k y+h^{2}+k^{2}-r^{2}=0\;\;\;\;\;\;\;\;\;\;\;\ldots(i)} \\ {\text { Which is of the form : }} \\ {\mathbf{x}^{2}+\mathbf{y}^{2}+2 \mathbf{g} \mathbf{x}+2 \mathbf{f} \mathbf{y}+\mathbf{c}=\mathbf{0}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(ii)}

This is known as the general equation of the circle.

Compare eq (i) and eq (ii)

h = -g, k = -h   and c=h2+k2-r2

Coordinates of the centre  (-g,-f)

Radius =g2+f2-c  


\\Re\left ( \frac{z-1}{2z+i} \right )=1


\operatorname{Re}\left(\frac{(x+i y)-1}{2(x+i y)+i}\right)=1

\operatorname{Re}\left(\left(\frac{(x-1)+i y}{2 x+i(2 y+1)}\right)\left(\frac{2 x-i(2 y+1}{2 x-i(2 y+1)}\right)\right)=1

\Rightarrow 2 x^{2}+2 y^{2}+2 x+3 y+1=0

x^{2}+y^{2}+x+\frac{3}{2} y+\frac{1}{2}=0

\Rightarrow $ locus is a circle whose

\text {Centre is }\left(-\frac{1}{2},-\frac{3}{4}\right)\text { and radius } \frac{\sqrt{5}}{4}.

\Rightarrow \text { diameter }=\frac{\sqrt{5}}{2}

Correct Option (3)

Posted by

Ritika Jonwal

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