117886
The equation \(|\mathbf{z}-\mathbf{i}|=|\mathbf{z}-\mathbf{1}|, \mathbf{i}=\sqrt{-1}\), represents
1 a circle of radius \(\frac{1}{2}\)
2 line passing through the origin with slope 1
3 a circle of radius 1
4 line passing through the origin with slope -1
Explanation:
B Given, \(z=x+i y\) \(|z-i|=|z-1|\) \(|\mathrm{x}+\mathrm{iy}-\mathrm{i}|=|\mathrm{x}+\mathrm{iy}-1|\) \(\mathrm{x}^2+(\mathrm{y}-1)^2=(\mathrm{x}-1)^2+\mathrm{y}^2\) \(\mathrm{x}^2+\mathrm{y}^2+1-2 \mathrm{y}=\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2\) \(\mathrm{x}=\mathrm{y}\) which is a equation of line passing through the origin with slope 1 .
JEE Main 12.04.2019
Complex Numbers and Quadratic Equation
117823
If \((\sqrt{5}+\sqrt{3} i)^{33}=2^{49} z\), then modulus of the complex number \(\mathrm{z}\) is equal to
1 1
2 \(\sqrt{2}\)
3 \(2 \sqrt{2}\)
4 4
Explanation:
B Given, \((\sqrt{5}+\sqrt{3 i})^{33}=2^{49} \mathrm{z}\) Taking modulus on both sides, we get \(2^{49}|z|=|\sqrt{5}+\sqrt{3} \mathrm{i}|^{33}\) \(2^{49}|z|=(\sqrt{5+3})^{33}=(\sqrt{8})^{33}=2^{33} 2^{33 / 2}\) \(|z|=\frac{2^{33} 2^{33 / 2}}{2^{49}}=2^{\frac{33}{2}-16}=2^{1 / 2}=\sqrt{2} right.\)
COMEDK-2017
Complex Numbers and Quadratic Equation
117810
If \(x^3-6 x^2+12 x+19=0\) and \(\omega\) is a non-real cube root of 1 , then \(x=\)
1 -1
2 \(2-3 \omega\)
3 \(2-3 \omega^2\)
4 (a) or (b) (c)
Explanation:
D Given that, \(\mathrm{x}^3-6 \mathrm{x}^2+12 \mathrm{x}+19=0\) Let put the value of \(x=-1\) in above equation Then, \(\quad(-1)^3-6(-1)^2+12(-1)+19=0\) \(\therefore \quad \mathrm{x}=-1\) is the root of equation \(\therefore \quad(\mathrm{x}+1)\left(\mathrm{x}^2-7 \mathrm{x}+19\right)=0\) \(\mathrm{x}^2-7 \mathrm{x}+19=0\) \(x=\frac{+7 \pm \sqrt{49-76}}{2}=\frac{7 \pm \sqrt{27 i^2}}{2}\) \(x=\frac{7 \pm i \sqrt{27}}{2}\) \(x=\frac{7+i \sqrt{27}}{2}, \frac{7-i \sqrt{27}}{2}\) We know, that \(\omega^2=-1-\frac{i \sqrt{3}}{2}, \omega =\frac{-1+i \sqrt{3}}{2}\) \(\therefore \quad \frac{7+i 3 \sqrt{3}}{2} =2-3 \omega^2\) \(\frac{7-i 3 \sqrt{3}}{2} =2-3 \omega\)Hence, \(x=-1,2-3 \omega, 2-3 \omega^2\)
SRM JEEE 2018
Complex Numbers and Quadratic Equation
117811
If \(\alpha\) is an \(n^{\text {th }}\) root of unity, other 1 , then \(1+2 \alpha\) \(+3 \alpha^2+\ldots \ldots+n \cdot \alpha^{n-1}\) equals
1 \(\frac{\mathrm{n}}{1-\alpha}\)
2 \(-\frac{\mathrm{n}}{1-\alpha}\)
3 \(-\frac{\mathrm{n}}{(1-\alpha)^2}\)
4 none of these
Explanation:
B \( Let \mathrm{S}=1+2 \alpha+3 \alpha^2+\ldots . .+\mathrm{n} \alpha^{\mathrm{n}-1} \ldots . . \text { (i) }\) \(\alpha \mathrm{S}=\alpha+2 \alpha^2+\ldots \ldots .+\mathrm{n} \alpha^{\mathrm{n}} \text {......(ii) }\) \(\mathrm{S}(1-\alpha)=\left[1+\alpha+\alpha^2+\alpha^3+\ldots . . \text { to } n \text { terms }\right]-n \alpha^{\mathrm{n}}\) \(\mathrm{S}(1-\alpha)=\frac{1-\alpha^{\mathrm{n}}}{1-\alpha}-\mathrm{n}^{\mathrm{n}}\) \(\mathrm{S}(1-\alpha)=\frac{1-1}{1-\alpha}-\mathrm{n} \quad\left[\because \alpha^{\mathrm{n}}=1\right]\) \(\mathrm{S}=\frac{-\mathrm{n}}{1-\alpha}\)Subtract equation (ii) from equation (i)
SRM JEEE-2008
Complex Numbers and Quadratic Equation
117812
If \(\mathrm{z}_{\mathrm{r}}=\cos \frac{21 \pi}{5}+\mathrm{i} \sin \frac{21 \pi}{5}, \mathrm{r}=0,1,2,3,4\), then \(\mathrm{z}_0\) \(z_1 z_2 z_3 z_4\) is equal to
117886
The equation \(|\mathbf{z}-\mathbf{i}|=|\mathbf{z}-\mathbf{1}|, \mathbf{i}=\sqrt{-1}\), represents
1 a circle of radius \(\frac{1}{2}\)
2 line passing through the origin with slope 1
3 a circle of radius 1
4 line passing through the origin with slope -1
Explanation:
B Given, \(z=x+i y\) \(|z-i|=|z-1|\) \(|\mathrm{x}+\mathrm{iy}-\mathrm{i}|=|\mathrm{x}+\mathrm{iy}-1|\) \(\mathrm{x}^2+(\mathrm{y}-1)^2=(\mathrm{x}-1)^2+\mathrm{y}^2\) \(\mathrm{x}^2+\mathrm{y}^2+1-2 \mathrm{y}=\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2\) \(\mathrm{x}=\mathrm{y}\) which is a equation of line passing through the origin with slope 1 .
JEE Main 12.04.2019
Complex Numbers and Quadratic Equation
117823
If \((\sqrt{5}+\sqrt{3} i)^{33}=2^{49} z\), then modulus of the complex number \(\mathrm{z}\) is equal to
1 1
2 \(\sqrt{2}\)
3 \(2 \sqrt{2}\)
4 4
Explanation:
B Given, \((\sqrt{5}+\sqrt{3 i})^{33}=2^{49} \mathrm{z}\) Taking modulus on both sides, we get \(2^{49}|z|=|\sqrt{5}+\sqrt{3} \mathrm{i}|^{33}\) \(2^{49}|z|=(\sqrt{5+3})^{33}=(\sqrt{8})^{33}=2^{33} 2^{33 / 2}\) \(|z|=\frac{2^{33} 2^{33 / 2}}{2^{49}}=2^{\frac{33}{2}-16}=2^{1 / 2}=\sqrt{2} right.\)
COMEDK-2017
Complex Numbers and Quadratic Equation
117810
If \(x^3-6 x^2+12 x+19=0\) and \(\omega\) is a non-real cube root of 1 , then \(x=\)
1 -1
2 \(2-3 \omega\)
3 \(2-3 \omega^2\)
4 (a) or (b) (c)
Explanation:
D Given that, \(\mathrm{x}^3-6 \mathrm{x}^2+12 \mathrm{x}+19=0\) Let put the value of \(x=-1\) in above equation Then, \(\quad(-1)^3-6(-1)^2+12(-1)+19=0\) \(\therefore \quad \mathrm{x}=-1\) is the root of equation \(\therefore \quad(\mathrm{x}+1)\left(\mathrm{x}^2-7 \mathrm{x}+19\right)=0\) \(\mathrm{x}^2-7 \mathrm{x}+19=0\) \(x=\frac{+7 \pm \sqrt{49-76}}{2}=\frac{7 \pm \sqrt{27 i^2}}{2}\) \(x=\frac{7 \pm i \sqrt{27}}{2}\) \(x=\frac{7+i \sqrt{27}}{2}, \frac{7-i \sqrt{27}}{2}\) We know, that \(\omega^2=-1-\frac{i \sqrt{3}}{2}, \omega =\frac{-1+i \sqrt{3}}{2}\) \(\therefore \quad \frac{7+i 3 \sqrt{3}}{2} =2-3 \omega^2\) \(\frac{7-i 3 \sqrt{3}}{2} =2-3 \omega\)Hence, \(x=-1,2-3 \omega, 2-3 \omega^2\)
SRM JEEE 2018
Complex Numbers and Quadratic Equation
117811
If \(\alpha\) is an \(n^{\text {th }}\) root of unity, other 1 , then \(1+2 \alpha\) \(+3 \alpha^2+\ldots \ldots+n \cdot \alpha^{n-1}\) equals
1 \(\frac{\mathrm{n}}{1-\alpha}\)
2 \(-\frac{\mathrm{n}}{1-\alpha}\)
3 \(-\frac{\mathrm{n}}{(1-\alpha)^2}\)
4 none of these
Explanation:
B \( Let \mathrm{S}=1+2 \alpha+3 \alpha^2+\ldots . .+\mathrm{n} \alpha^{\mathrm{n}-1} \ldots . . \text { (i) }\) \(\alpha \mathrm{S}=\alpha+2 \alpha^2+\ldots \ldots .+\mathrm{n} \alpha^{\mathrm{n}} \text {......(ii) }\) \(\mathrm{S}(1-\alpha)=\left[1+\alpha+\alpha^2+\alpha^3+\ldots . . \text { to } n \text { terms }\right]-n \alpha^{\mathrm{n}}\) \(\mathrm{S}(1-\alpha)=\frac{1-\alpha^{\mathrm{n}}}{1-\alpha}-\mathrm{n}^{\mathrm{n}}\) \(\mathrm{S}(1-\alpha)=\frac{1-1}{1-\alpha}-\mathrm{n} \quad\left[\because \alpha^{\mathrm{n}}=1\right]\) \(\mathrm{S}=\frac{-\mathrm{n}}{1-\alpha}\)Subtract equation (ii) from equation (i)
SRM JEEE-2008
Complex Numbers and Quadratic Equation
117812
If \(\mathrm{z}_{\mathrm{r}}=\cos \frac{21 \pi}{5}+\mathrm{i} \sin \frac{21 \pi}{5}, \mathrm{r}=0,1,2,3,4\), then \(\mathrm{z}_0\) \(z_1 z_2 z_3 z_4\) is equal to
117886
The equation \(|\mathbf{z}-\mathbf{i}|=|\mathbf{z}-\mathbf{1}|, \mathbf{i}=\sqrt{-1}\), represents
1 a circle of radius \(\frac{1}{2}\)
2 line passing through the origin with slope 1
3 a circle of radius 1
4 line passing through the origin with slope -1
Explanation:
B Given, \(z=x+i y\) \(|z-i|=|z-1|\) \(|\mathrm{x}+\mathrm{iy}-\mathrm{i}|=|\mathrm{x}+\mathrm{iy}-1|\) \(\mathrm{x}^2+(\mathrm{y}-1)^2=(\mathrm{x}-1)^2+\mathrm{y}^2\) \(\mathrm{x}^2+\mathrm{y}^2+1-2 \mathrm{y}=\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2\) \(\mathrm{x}=\mathrm{y}\) which is a equation of line passing through the origin with slope 1 .
JEE Main 12.04.2019
Complex Numbers and Quadratic Equation
117823
If \((\sqrt{5}+\sqrt{3} i)^{33}=2^{49} z\), then modulus of the complex number \(\mathrm{z}\) is equal to
1 1
2 \(\sqrt{2}\)
3 \(2 \sqrt{2}\)
4 4
Explanation:
B Given, \((\sqrt{5}+\sqrt{3 i})^{33}=2^{49} \mathrm{z}\) Taking modulus on both sides, we get \(2^{49}|z|=|\sqrt{5}+\sqrt{3} \mathrm{i}|^{33}\) \(2^{49}|z|=(\sqrt{5+3})^{33}=(\sqrt{8})^{33}=2^{33} 2^{33 / 2}\) \(|z|=\frac{2^{33} 2^{33 / 2}}{2^{49}}=2^{\frac{33}{2}-16}=2^{1 / 2}=\sqrt{2} right.\)
COMEDK-2017
Complex Numbers and Quadratic Equation
117810
If \(x^3-6 x^2+12 x+19=0\) and \(\omega\) is a non-real cube root of 1 , then \(x=\)
1 -1
2 \(2-3 \omega\)
3 \(2-3 \omega^2\)
4 (a) or (b) (c)
Explanation:
D Given that, \(\mathrm{x}^3-6 \mathrm{x}^2+12 \mathrm{x}+19=0\) Let put the value of \(x=-1\) in above equation Then, \(\quad(-1)^3-6(-1)^2+12(-1)+19=0\) \(\therefore \quad \mathrm{x}=-1\) is the root of equation \(\therefore \quad(\mathrm{x}+1)\left(\mathrm{x}^2-7 \mathrm{x}+19\right)=0\) \(\mathrm{x}^2-7 \mathrm{x}+19=0\) \(x=\frac{+7 \pm \sqrt{49-76}}{2}=\frac{7 \pm \sqrt{27 i^2}}{2}\) \(x=\frac{7 \pm i \sqrt{27}}{2}\) \(x=\frac{7+i \sqrt{27}}{2}, \frac{7-i \sqrt{27}}{2}\) We know, that \(\omega^2=-1-\frac{i \sqrt{3}}{2}, \omega =\frac{-1+i \sqrt{3}}{2}\) \(\therefore \quad \frac{7+i 3 \sqrt{3}}{2} =2-3 \omega^2\) \(\frac{7-i 3 \sqrt{3}}{2} =2-3 \omega\)Hence, \(x=-1,2-3 \omega, 2-3 \omega^2\)
SRM JEEE 2018
Complex Numbers and Quadratic Equation
117811
If \(\alpha\) is an \(n^{\text {th }}\) root of unity, other 1 , then \(1+2 \alpha\) \(+3 \alpha^2+\ldots \ldots+n \cdot \alpha^{n-1}\) equals
1 \(\frac{\mathrm{n}}{1-\alpha}\)
2 \(-\frac{\mathrm{n}}{1-\alpha}\)
3 \(-\frac{\mathrm{n}}{(1-\alpha)^2}\)
4 none of these
Explanation:
B \( Let \mathrm{S}=1+2 \alpha+3 \alpha^2+\ldots . .+\mathrm{n} \alpha^{\mathrm{n}-1} \ldots . . \text { (i) }\) \(\alpha \mathrm{S}=\alpha+2 \alpha^2+\ldots \ldots .+\mathrm{n} \alpha^{\mathrm{n}} \text {......(ii) }\) \(\mathrm{S}(1-\alpha)=\left[1+\alpha+\alpha^2+\alpha^3+\ldots . . \text { to } n \text { terms }\right]-n \alpha^{\mathrm{n}}\) \(\mathrm{S}(1-\alpha)=\frac{1-\alpha^{\mathrm{n}}}{1-\alpha}-\mathrm{n}^{\mathrm{n}}\) \(\mathrm{S}(1-\alpha)=\frac{1-1}{1-\alpha}-\mathrm{n} \quad\left[\because \alpha^{\mathrm{n}}=1\right]\) \(\mathrm{S}=\frac{-\mathrm{n}}{1-\alpha}\)Subtract equation (ii) from equation (i)
SRM JEEE-2008
Complex Numbers and Quadratic Equation
117812
If \(\mathrm{z}_{\mathrm{r}}=\cos \frac{21 \pi}{5}+\mathrm{i} \sin \frac{21 \pi}{5}, \mathrm{r}=0,1,2,3,4\), then \(\mathrm{z}_0\) \(z_1 z_2 z_3 z_4\) is equal to
117886
The equation \(|\mathbf{z}-\mathbf{i}|=|\mathbf{z}-\mathbf{1}|, \mathbf{i}=\sqrt{-1}\), represents
1 a circle of radius \(\frac{1}{2}\)
2 line passing through the origin with slope 1
3 a circle of radius 1
4 line passing through the origin with slope -1
Explanation:
B Given, \(z=x+i y\) \(|z-i|=|z-1|\) \(|\mathrm{x}+\mathrm{iy}-\mathrm{i}|=|\mathrm{x}+\mathrm{iy}-1|\) \(\mathrm{x}^2+(\mathrm{y}-1)^2=(\mathrm{x}-1)^2+\mathrm{y}^2\) \(\mathrm{x}^2+\mathrm{y}^2+1-2 \mathrm{y}=\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2\) \(\mathrm{x}=\mathrm{y}\) which is a equation of line passing through the origin with slope 1 .
JEE Main 12.04.2019
Complex Numbers and Quadratic Equation
117823
If \((\sqrt{5}+\sqrt{3} i)^{33}=2^{49} z\), then modulus of the complex number \(\mathrm{z}\) is equal to
1 1
2 \(\sqrt{2}\)
3 \(2 \sqrt{2}\)
4 4
Explanation:
B Given, \((\sqrt{5}+\sqrt{3 i})^{33}=2^{49} \mathrm{z}\) Taking modulus on both sides, we get \(2^{49}|z|=|\sqrt{5}+\sqrt{3} \mathrm{i}|^{33}\) \(2^{49}|z|=(\sqrt{5+3})^{33}=(\sqrt{8})^{33}=2^{33} 2^{33 / 2}\) \(|z|=\frac{2^{33} 2^{33 / 2}}{2^{49}}=2^{\frac{33}{2}-16}=2^{1 / 2}=\sqrt{2} right.\)
COMEDK-2017
Complex Numbers and Quadratic Equation
117810
If \(x^3-6 x^2+12 x+19=0\) and \(\omega\) is a non-real cube root of 1 , then \(x=\)
1 -1
2 \(2-3 \omega\)
3 \(2-3 \omega^2\)
4 (a) or (b) (c)
Explanation:
D Given that, \(\mathrm{x}^3-6 \mathrm{x}^2+12 \mathrm{x}+19=0\) Let put the value of \(x=-1\) in above equation Then, \(\quad(-1)^3-6(-1)^2+12(-1)+19=0\) \(\therefore \quad \mathrm{x}=-1\) is the root of equation \(\therefore \quad(\mathrm{x}+1)\left(\mathrm{x}^2-7 \mathrm{x}+19\right)=0\) \(\mathrm{x}^2-7 \mathrm{x}+19=0\) \(x=\frac{+7 \pm \sqrt{49-76}}{2}=\frac{7 \pm \sqrt{27 i^2}}{2}\) \(x=\frac{7 \pm i \sqrt{27}}{2}\) \(x=\frac{7+i \sqrt{27}}{2}, \frac{7-i \sqrt{27}}{2}\) We know, that \(\omega^2=-1-\frac{i \sqrt{3}}{2}, \omega =\frac{-1+i \sqrt{3}}{2}\) \(\therefore \quad \frac{7+i 3 \sqrt{3}}{2} =2-3 \omega^2\) \(\frac{7-i 3 \sqrt{3}}{2} =2-3 \omega\)Hence, \(x=-1,2-3 \omega, 2-3 \omega^2\)
SRM JEEE 2018
Complex Numbers and Quadratic Equation
117811
If \(\alpha\) is an \(n^{\text {th }}\) root of unity, other 1 , then \(1+2 \alpha\) \(+3 \alpha^2+\ldots \ldots+n \cdot \alpha^{n-1}\) equals
1 \(\frac{\mathrm{n}}{1-\alpha}\)
2 \(-\frac{\mathrm{n}}{1-\alpha}\)
3 \(-\frac{\mathrm{n}}{(1-\alpha)^2}\)
4 none of these
Explanation:
B \( Let \mathrm{S}=1+2 \alpha+3 \alpha^2+\ldots . .+\mathrm{n} \alpha^{\mathrm{n}-1} \ldots . . \text { (i) }\) \(\alpha \mathrm{S}=\alpha+2 \alpha^2+\ldots \ldots .+\mathrm{n} \alpha^{\mathrm{n}} \text {......(ii) }\) \(\mathrm{S}(1-\alpha)=\left[1+\alpha+\alpha^2+\alpha^3+\ldots . . \text { to } n \text { terms }\right]-n \alpha^{\mathrm{n}}\) \(\mathrm{S}(1-\alpha)=\frac{1-\alpha^{\mathrm{n}}}{1-\alpha}-\mathrm{n}^{\mathrm{n}}\) \(\mathrm{S}(1-\alpha)=\frac{1-1}{1-\alpha}-\mathrm{n} \quad\left[\because \alpha^{\mathrm{n}}=1\right]\) \(\mathrm{S}=\frac{-\mathrm{n}}{1-\alpha}\)Subtract equation (ii) from equation (i)
SRM JEEE-2008
Complex Numbers and Quadratic Equation
117812
If \(\mathrm{z}_{\mathrm{r}}=\cos \frac{21 \pi}{5}+\mathrm{i} \sin \frac{21 \pi}{5}, \mathrm{r}=0,1,2,3,4\), then \(\mathrm{z}_0\) \(z_1 z_2 z_3 z_4\) is equal to
117886
The equation \(|\mathbf{z}-\mathbf{i}|=|\mathbf{z}-\mathbf{1}|, \mathbf{i}=\sqrt{-1}\), represents
1 a circle of radius \(\frac{1}{2}\)
2 line passing through the origin with slope 1
3 a circle of radius 1
4 line passing through the origin with slope -1
Explanation:
B Given, \(z=x+i y\) \(|z-i|=|z-1|\) \(|\mathrm{x}+\mathrm{iy}-\mathrm{i}|=|\mathrm{x}+\mathrm{iy}-1|\) \(\mathrm{x}^2+(\mathrm{y}-1)^2=(\mathrm{x}-1)^2+\mathrm{y}^2\) \(\mathrm{x}^2+\mathrm{y}^2+1-2 \mathrm{y}=\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2\) \(\mathrm{x}=\mathrm{y}\) which is a equation of line passing through the origin with slope 1 .
JEE Main 12.04.2019
Complex Numbers and Quadratic Equation
117823
If \((\sqrt{5}+\sqrt{3} i)^{33}=2^{49} z\), then modulus of the complex number \(\mathrm{z}\) is equal to
1 1
2 \(\sqrt{2}\)
3 \(2 \sqrt{2}\)
4 4
Explanation:
B Given, \((\sqrt{5}+\sqrt{3 i})^{33}=2^{49} \mathrm{z}\) Taking modulus on both sides, we get \(2^{49}|z|=|\sqrt{5}+\sqrt{3} \mathrm{i}|^{33}\) \(2^{49}|z|=(\sqrt{5+3})^{33}=(\sqrt{8})^{33}=2^{33} 2^{33 / 2}\) \(|z|=\frac{2^{33} 2^{33 / 2}}{2^{49}}=2^{\frac{33}{2}-16}=2^{1 / 2}=\sqrt{2} right.\)
COMEDK-2017
Complex Numbers and Quadratic Equation
117810
If \(x^3-6 x^2+12 x+19=0\) and \(\omega\) is a non-real cube root of 1 , then \(x=\)
1 -1
2 \(2-3 \omega\)
3 \(2-3 \omega^2\)
4 (a) or (b) (c)
Explanation:
D Given that, \(\mathrm{x}^3-6 \mathrm{x}^2+12 \mathrm{x}+19=0\) Let put the value of \(x=-1\) in above equation Then, \(\quad(-1)^3-6(-1)^2+12(-1)+19=0\) \(\therefore \quad \mathrm{x}=-1\) is the root of equation \(\therefore \quad(\mathrm{x}+1)\left(\mathrm{x}^2-7 \mathrm{x}+19\right)=0\) \(\mathrm{x}^2-7 \mathrm{x}+19=0\) \(x=\frac{+7 \pm \sqrt{49-76}}{2}=\frac{7 \pm \sqrt{27 i^2}}{2}\) \(x=\frac{7 \pm i \sqrt{27}}{2}\) \(x=\frac{7+i \sqrt{27}}{2}, \frac{7-i \sqrt{27}}{2}\) We know, that \(\omega^2=-1-\frac{i \sqrt{3}}{2}, \omega =\frac{-1+i \sqrt{3}}{2}\) \(\therefore \quad \frac{7+i 3 \sqrt{3}}{2} =2-3 \omega^2\) \(\frac{7-i 3 \sqrt{3}}{2} =2-3 \omega\)Hence, \(x=-1,2-3 \omega, 2-3 \omega^2\)
SRM JEEE 2018
Complex Numbers and Quadratic Equation
117811
If \(\alpha\) is an \(n^{\text {th }}\) root of unity, other 1 , then \(1+2 \alpha\) \(+3 \alpha^2+\ldots \ldots+n \cdot \alpha^{n-1}\) equals
1 \(\frac{\mathrm{n}}{1-\alpha}\)
2 \(-\frac{\mathrm{n}}{1-\alpha}\)
3 \(-\frac{\mathrm{n}}{(1-\alpha)^2}\)
4 none of these
Explanation:
B \( Let \mathrm{S}=1+2 \alpha+3 \alpha^2+\ldots . .+\mathrm{n} \alpha^{\mathrm{n}-1} \ldots . . \text { (i) }\) \(\alpha \mathrm{S}=\alpha+2 \alpha^2+\ldots \ldots .+\mathrm{n} \alpha^{\mathrm{n}} \text {......(ii) }\) \(\mathrm{S}(1-\alpha)=\left[1+\alpha+\alpha^2+\alpha^3+\ldots . . \text { to } n \text { terms }\right]-n \alpha^{\mathrm{n}}\) \(\mathrm{S}(1-\alpha)=\frac{1-\alpha^{\mathrm{n}}}{1-\alpha}-\mathrm{n}^{\mathrm{n}}\) \(\mathrm{S}(1-\alpha)=\frac{1-1}{1-\alpha}-\mathrm{n} \quad\left[\because \alpha^{\mathrm{n}}=1\right]\) \(\mathrm{S}=\frac{-\mathrm{n}}{1-\alpha}\)Subtract equation (ii) from equation (i)
SRM JEEE-2008
Complex Numbers and Quadratic Equation
117812
If \(\mathrm{z}_{\mathrm{r}}=\cos \frac{21 \pi}{5}+\mathrm{i} \sin \frac{21 \pi}{5}, \mathrm{r}=0,1,2,3,4\), then \(\mathrm{z}_0\) \(z_1 z_2 z_3 z_4\) is equal to
