117972
If the real part of \(\frac{\bar{z}+2}{\bar{z}-1}\) is \(4, z \neq 1\), then the locus of the point representing \(\mathrm{z}\) in the complex plane is
1 a straight line parallel to \(x\)-axis
2 a straight equally inclined to axes
3 a circle with radius 2
4 a circle with radius \(\frac{1}{2}\)
Explanation:
D Let, \(\mathrm{z}=\mathrm{x}+\) iy Now, \(\frac{\bar{z}+2}{\bar{z}-2}=\frac{x-i y+2}{x-i y-1}\) \(=\frac{[(x+2)-i y][(x-1)+i y]}{[(x-1)-i y][(x-1)+i y]}\) \(=\frac{(x-1)(x+2)+y^2+i[(x+2) y-(x-1) y]}{(x-1)^2+y^2}\) Given that real part is 4 . \(\frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4\) \(x^2+x-2+y^2=4\left(x^2-2 x+1+y^2\right)\) \(3 x^2+3 y^2-9 x+6=0\) \(x^2+y^2-3 x+2=0\) This equation represents a circle. and, the radius are \(\mathrm{r} =\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{\left(\frac{-3}{2}\right)^2+0^2-2}\) \(=\sqrt{\frac{9}{4}-2}=\frac{1}{2}\)
BITSAT-2013
Complex Numbers and Quadratic Equation
117973
If \(\omega\) is the cubic root of unity, then value of the \(\left(1+\omega-\omega^2\right)^2+\left(1-\omega+\omega^2\right)^2+1\) is
1 -1
2 7
3 1
4 -3
Explanation:
D \(\left(1+\omega-\omega^2\right)^2+\left(1-\omega+\omega^2\right)^2+1\) \(\because \omega\) is cube root of unity \(\therefore 1+\omega +\omega^2=0\) \(=\left(-\omega^2-\omega^2\right)^2+(-\omega-\omega)^2+1\) \(=4 \omega^4+4 \omega^2+1\) \(=4 \omega+4 \omega^2+1\) \(=4\left(\omega+\omega^2\right)+1\) \(=4 \times(-1)+1=-4+1=-3\)
UPSEE-2016
Complex Numbers and Quadratic Equation
117974
For two complex numbers \(z_1, z_2\) the relation \(\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|\) hold, if
117972
If the real part of \(\frac{\bar{z}+2}{\bar{z}-1}\) is \(4, z \neq 1\), then the locus of the point representing \(\mathrm{z}\) in the complex plane is
1 a straight line parallel to \(x\)-axis
2 a straight equally inclined to axes
3 a circle with radius 2
4 a circle with radius \(\frac{1}{2}\)
Explanation:
D Let, \(\mathrm{z}=\mathrm{x}+\) iy Now, \(\frac{\bar{z}+2}{\bar{z}-2}=\frac{x-i y+2}{x-i y-1}\) \(=\frac{[(x+2)-i y][(x-1)+i y]}{[(x-1)-i y][(x-1)+i y]}\) \(=\frac{(x-1)(x+2)+y^2+i[(x+2) y-(x-1) y]}{(x-1)^2+y^2}\) Given that real part is 4 . \(\frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4\) \(x^2+x-2+y^2=4\left(x^2-2 x+1+y^2\right)\) \(3 x^2+3 y^2-9 x+6=0\) \(x^2+y^2-3 x+2=0\) This equation represents a circle. and, the radius are \(\mathrm{r} =\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{\left(\frac{-3}{2}\right)^2+0^2-2}\) \(=\sqrt{\frac{9}{4}-2}=\frac{1}{2}\)
BITSAT-2013
Complex Numbers and Quadratic Equation
117973
If \(\omega\) is the cubic root of unity, then value of the \(\left(1+\omega-\omega^2\right)^2+\left(1-\omega+\omega^2\right)^2+1\) is
1 -1
2 7
3 1
4 -3
Explanation:
D \(\left(1+\omega-\omega^2\right)^2+\left(1-\omega+\omega^2\right)^2+1\) \(\because \omega\) is cube root of unity \(\therefore 1+\omega +\omega^2=0\) \(=\left(-\omega^2-\omega^2\right)^2+(-\omega-\omega)^2+1\) \(=4 \omega^4+4 \omega^2+1\) \(=4 \omega+4 \omega^2+1\) \(=4\left(\omega+\omega^2\right)+1\) \(=4 \times(-1)+1=-4+1=-3\)
UPSEE-2016
Complex Numbers and Quadratic Equation
117974
For two complex numbers \(z_1, z_2\) the relation \(\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|\) hold, if
117972
If the real part of \(\frac{\bar{z}+2}{\bar{z}-1}\) is \(4, z \neq 1\), then the locus of the point representing \(\mathrm{z}\) in the complex plane is
1 a straight line parallel to \(x\)-axis
2 a straight equally inclined to axes
3 a circle with radius 2
4 a circle with radius \(\frac{1}{2}\)
Explanation:
D Let, \(\mathrm{z}=\mathrm{x}+\) iy Now, \(\frac{\bar{z}+2}{\bar{z}-2}=\frac{x-i y+2}{x-i y-1}\) \(=\frac{[(x+2)-i y][(x-1)+i y]}{[(x-1)-i y][(x-1)+i y]}\) \(=\frac{(x-1)(x+2)+y^2+i[(x+2) y-(x-1) y]}{(x-1)^2+y^2}\) Given that real part is 4 . \(\frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4\) \(x^2+x-2+y^2=4\left(x^2-2 x+1+y^2\right)\) \(3 x^2+3 y^2-9 x+6=0\) \(x^2+y^2-3 x+2=0\) This equation represents a circle. and, the radius are \(\mathrm{r} =\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{\left(\frac{-3}{2}\right)^2+0^2-2}\) \(=\sqrt{\frac{9}{4}-2}=\frac{1}{2}\)
BITSAT-2013
Complex Numbers and Quadratic Equation
117973
If \(\omega\) is the cubic root of unity, then value of the \(\left(1+\omega-\omega^2\right)^2+\left(1-\omega+\omega^2\right)^2+1\) is
1 -1
2 7
3 1
4 -3
Explanation:
D \(\left(1+\omega-\omega^2\right)^2+\left(1-\omega+\omega^2\right)^2+1\) \(\because \omega\) is cube root of unity \(\therefore 1+\omega +\omega^2=0\) \(=\left(-\omega^2-\omega^2\right)^2+(-\omega-\omega)^2+1\) \(=4 \omega^4+4 \omega^2+1\) \(=4 \omega+4 \omega^2+1\) \(=4\left(\omega+\omega^2\right)+1\) \(=4 \times(-1)+1=-4+1=-3\)
UPSEE-2016
Complex Numbers and Quadratic Equation
117974
For two complex numbers \(z_1, z_2\) the relation \(\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|\) hold, if
117972
If the real part of \(\frac{\bar{z}+2}{\bar{z}-1}\) is \(4, z \neq 1\), then the locus of the point representing \(\mathrm{z}\) in the complex plane is
1 a straight line parallel to \(x\)-axis
2 a straight equally inclined to axes
3 a circle with radius 2
4 a circle with radius \(\frac{1}{2}\)
Explanation:
D Let, \(\mathrm{z}=\mathrm{x}+\) iy Now, \(\frac{\bar{z}+2}{\bar{z}-2}=\frac{x-i y+2}{x-i y-1}\) \(=\frac{[(x+2)-i y][(x-1)+i y]}{[(x-1)-i y][(x-1)+i y]}\) \(=\frac{(x-1)(x+2)+y^2+i[(x+2) y-(x-1) y]}{(x-1)^2+y^2}\) Given that real part is 4 . \(\frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4\) \(x^2+x-2+y^2=4\left(x^2-2 x+1+y^2\right)\) \(3 x^2+3 y^2-9 x+6=0\) \(x^2+y^2-3 x+2=0\) This equation represents a circle. and, the radius are \(\mathrm{r} =\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{\left(\frac{-3}{2}\right)^2+0^2-2}\) \(=\sqrt{\frac{9}{4}-2}=\frac{1}{2}\)
BITSAT-2013
Complex Numbers and Quadratic Equation
117973
If \(\omega\) is the cubic root of unity, then value of the \(\left(1+\omega-\omega^2\right)^2+\left(1-\omega+\omega^2\right)^2+1\) is
1 -1
2 7
3 1
4 -3
Explanation:
D \(\left(1+\omega-\omega^2\right)^2+\left(1-\omega+\omega^2\right)^2+1\) \(\because \omega\) is cube root of unity \(\therefore 1+\omega +\omega^2=0\) \(=\left(-\omega^2-\omega^2\right)^2+(-\omega-\omega)^2+1\) \(=4 \omega^4+4 \omega^2+1\) \(=4 \omega+4 \omega^2+1\) \(=4\left(\omega+\omega^2\right)+1\) \(=4 \times(-1)+1=-4+1=-3\)
UPSEE-2016
Complex Numbers and Quadratic Equation
117974
For two complex numbers \(z_1, z_2\) the relation \(\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|\) hold, if
117972
If the real part of \(\frac{\bar{z}+2}{\bar{z}-1}\) is \(4, z \neq 1\), then the locus of the point representing \(\mathrm{z}\) in the complex plane is
1 a straight line parallel to \(x\)-axis
2 a straight equally inclined to axes
3 a circle with radius 2
4 a circle with radius \(\frac{1}{2}\)
Explanation:
D Let, \(\mathrm{z}=\mathrm{x}+\) iy Now, \(\frac{\bar{z}+2}{\bar{z}-2}=\frac{x-i y+2}{x-i y-1}\) \(=\frac{[(x+2)-i y][(x-1)+i y]}{[(x-1)-i y][(x-1)+i y]}\) \(=\frac{(x-1)(x+2)+y^2+i[(x+2) y-(x-1) y]}{(x-1)^2+y^2}\) Given that real part is 4 . \(\frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4\) \(x^2+x-2+y^2=4\left(x^2-2 x+1+y^2\right)\) \(3 x^2+3 y^2-9 x+6=0\) \(x^2+y^2-3 x+2=0\) This equation represents a circle. and, the radius are \(\mathrm{r} =\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{\left(\frac{-3}{2}\right)^2+0^2-2}\) \(=\sqrt{\frac{9}{4}-2}=\frac{1}{2}\)
BITSAT-2013
Complex Numbers and Quadratic Equation
117973
If \(\omega\) is the cubic root of unity, then value of the \(\left(1+\omega-\omega^2\right)^2+\left(1-\omega+\omega^2\right)^2+1\) is
1 -1
2 7
3 1
4 -3
Explanation:
D \(\left(1+\omega-\omega^2\right)^2+\left(1-\omega+\omega^2\right)^2+1\) \(\because \omega\) is cube root of unity \(\therefore 1+\omega +\omega^2=0\) \(=\left(-\omega^2-\omega^2\right)^2+(-\omega-\omega)^2+1\) \(=4 \omega^4+4 \omega^2+1\) \(=4 \omega+4 \omega^2+1\) \(=4\left(\omega+\omega^2\right)+1\) \(=4 \times(-1)+1=-4+1=-3\)
UPSEE-2016
Complex Numbers and Quadratic Equation
117974
For two complex numbers \(z_1, z_2\) the relation \(\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|\) hold, if
