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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)

best_answer

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.

Note:

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

-

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

$\\\text{put}\;z=x+iy$

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

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

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

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

$\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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