117844
If \(\frac{|z-2|}{|z-3|}=2\) represents a circle, then its radius is equal to
1 1
2 \(\frac{1}{3}\)
3 \(\frac{3}{4}\)
4 \(\frac{2}{3}\)
Explanation:
D Given that, \(\frac{|z-2|}{|z-3|}=2\) Let, \(\quad z=x+i y\) Then, \(\quad \frac{|x+i y-2|}{|x+i y-3|}=2\) \(\mid(x-2)+\text { iy }|=2|(x-3)+i y \mid\) \(\sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2}\) Squaring both side - \((x-2)^2+y^2=4\left\{(x-3)^2+y^2\right\}\) \(x^2+4-4 x+y^2=4 x^2+36-24 x+4 y^2\) \(3 x^2+3 y^2-20 x+32=0\) \(x^2+y^2-\frac{20}{3} x+\frac{32}{3}=0\) Adding \(\frac{100}{9}\) both side \(\left(x-\frac{10}{3}\right)^2+y^2+\frac{32}{3}=\frac{100}{9}\) \(\left(x-\frac{10}{3}\right)^2+y^2=\frac{100}{9}-\frac{32}{3}\) \(\left(x-\frac{10}{3}\right)^2+y^2=\frac{4}{9}\)Above equation represent a circle with center \(\left(\frac{10}{3}, 0\right)\) and radius \(\frac{2}{3}\)
JCECE-2014
Complex Numbers and Quadratic Equation
117845
If \(\theta\) is real and \(z_1, z_2\) are connected by \(\mathrm{z}_1^2+\mathrm{z}_2^2+2 \mathrm{z}_1 \mathrm{z}_2 \cos \theta=0\), then triangle with vertices \(0, z_1\) and \(z_2\) is
1 equilateral
2 right angled
3 isosceles
4 None of above
Explanation:
C Given that, \(z_1^2+z_2^2+2 z_1 z_2 \cos \theta=0\) Now, above equation is divided by \(\left(z_2\right)^2\) \(\frac{z_1^2}{z_2^2}+1+\frac{2 z_1}{z_2} \cos \theta=0\) \(\left(\frac{z_1}{z_2}\right)^2+2\left(\frac{z_1}{z_2}\right) \cos \theta+\cos ^2 \theta+\sin ^2 \theta=0\) \(\left(\frac{z_1}{z_2}+\cos \theta\right)^2+\sin ^2 \theta=0\) \(\left.\left(\frac{z_1}{z_2}+\cos \theta\right)^2=-\sin ^2 \theta=1\right)\) \(\frac{z_1}{z_2}+\cos \theta= \pm \sin \theta\) \(\frac{z_1}{z_2}=-\cos \theta \pm \sin \theta\) \(\left|\frac{z_1}{z_2}\right|=\sqrt{(-\cos \theta)^2+\sin ^2 \theta}=\sqrt{1}=1\) \(\left|z_1\right|=\left|z_2\right|\) \(\left|z_1-0\right|=\left|z_2-0\right|\)From above we can say that, triangle with vertices \(0, z_1\) and \(z_2\) is isosceles.
JCECE-2013
Complex Numbers and Quadratic Equation
117846
If \(\mathrm{z}+\mathrm{z}^{-1}=1\), then \(\mathrm{z}^{100}+\mathrm{z}^{-100}\) is equal to
1 i
2 \(-\mathrm{i}\)
3 1
4 -1
Explanation:
D Given that, \(\mathrm{z}+\mathrm{z}^{-1}=1\) Then, \(\quad z+\frac{1}{z}=1\) \(\mathrm{z}^2+1-\mathrm{z}=0\) So, \(\quad z=\frac{1 \pm \sqrt{(1)^2-4}}{2}\) \(=\frac{1 \pm \sqrt{3} \mathrm{i}}{2}\) \(\mathrm{z} =\frac{1+\mathrm{i} \sqrt{3}}{2}, \frac{1-\mathrm{i} \sqrt{3}}{2}\) \(\mathrm{z} =-\left(\frac{-1-\mathrm{i} \sqrt{3}}{2}\right),-\left(\frac{-1+\mathrm{i} \sqrt{3}}{2}\right)\) \(z=-\omega^2,-\omega \quad\{\) where \(\omega\) is cube root of unity) Now, \(\quad z^{100}+\frac{1}{z^{100}}=(-\omega)^{100}+\frac{1}{(-\omega)^{100}}\) \(=\omega^{100}+\frac{1}{\omega^{100}}\) \(=\left(\omega^3\right)^{33} \cdot \omega+\) \(=\omega+\frac{1}{\omega}\) \(=\omega+\omega^2\) \(=-1\)
JCECE-2011
Complex Numbers and Quadratic Equation
117847
If \(x=a+b, y=a \omega+b \omega^2\) and \(z=a \omega^2+b \omega\), then \(x y z\) is equal to
1 \((a+b)^3\)
2 \(a^3+b^3\)
3 \(a^3-b^3\)
4 \((a+b)^3-3 a b(a+b)\)
Explanation:
B Given that, \(x=a+b, y=a \omega+b \omega^2 \text { and } z=a \omega^2+b \omega\) \(\text { then, } \quad x \cdot y \cdot z\) \(=(a+b)\left(a \omega+b \omega^2\right)\left(a \omega^2+b \omega\right)\) \(=(a+b)\left(a^2 \omega^3+a b \omega^2+a b \omega^4+b^2 \cdot \omega^3\right)\) \((a+b)\left(a^2+a b \omega^2+a b \omega+b^2\right)\) \((a+b)\left(a^2+b^2+a b\left(\omega^2+\omega\right)\right.\) \((a+b)\left(a^2+b^2-a b\right)\) \(=a^3+b^3\)
117844
If \(\frac{|z-2|}{|z-3|}=2\) represents a circle, then its radius is equal to
1 1
2 \(\frac{1}{3}\)
3 \(\frac{3}{4}\)
4 \(\frac{2}{3}\)
Explanation:
D Given that, \(\frac{|z-2|}{|z-3|}=2\) Let, \(\quad z=x+i y\) Then, \(\quad \frac{|x+i y-2|}{|x+i y-3|}=2\) \(\mid(x-2)+\text { iy }|=2|(x-3)+i y \mid\) \(\sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2}\) Squaring both side - \((x-2)^2+y^2=4\left\{(x-3)^2+y^2\right\}\) \(x^2+4-4 x+y^2=4 x^2+36-24 x+4 y^2\) \(3 x^2+3 y^2-20 x+32=0\) \(x^2+y^2-\frac{20}{3} x+\frac{32}{3}=0\) Adding \(\frac{100}{9}\) both side \(\left(x-\frac{10}{3}\right)^2+y^2+\frac{32}{3}=\frac{100}{9}\) \(\left(x-\frac{10}{3}\right)^2+y^2=\frac{100}{9}-\frac{32}{3}\) \(\left(x-\frac{10}{3}\right)^2+y^2=\frac{4}{9}\)Above equation represent a circle with center \(\left(\frac{10}{3}, 0\right)\) and radius \(\frac{2}{3}\)
JCECE-2014
Complex Numbers and Quadratic Equation
117845
If \(\theta\) is real and \(z_1, z_2\) are connected by \(\mathrm{z}_1^2+\mathrm{z}_2^2+2 \mathrm{z}_1 \mathrm{z}_2 \cos \theta=0\), then triangle with vertices \(0, z_1\) and \(z_2\) is
1 equilateral
2 right angled
3 isosceles
4 None of above
Explanation:
C Given that, \(z_1^2+z_2^2+2 z_1 z_2 \cos \theta=0\) Now, above equation is divided by \(\left(z_2\right)^2\) \(\frac{z_1^2}{z_2^2}+1+\frac{2 z_1}{z_2} \cos \theta=0\) \(\left(\frac{z_1}{z_2}\right)^2+2\left(\frac{z_1}{z_2}\right) \cos \theta+\cos ^2 \theta+\sin ^2 \theta=0\) \(\left(\frac{z_1}{z_2}+\cos \theta\right)^2+\sin ^2 \theta=0\) \(\left.\left(\frac{z_1}{z_2}+\cos \theta\right)^2=-\sin ^2 \theta=1\right)\) \(\frac{z_1}{z_2}+\cos \theta= \pm \sin \theta\) \(\frac{z_1}{z_2}=-\cos \theta \pm \sin \theta\) \(\left|\frac{z_1}{z_2}\right|=\sqrt{(-\cos \theta)^2+\sin ^2 \theta}=\sqrt{1}=1\) \(\left|z_1\right|=\left|z_2\right|\) \(\left|z_1-0\right|=\left|z_2-0\right|\)From above we can say that, triangle with vertices \(0, z_1\) and \(z_2\) is isosceles.
JCECE-2013
Complex Numbers and Quadratic Equation
117846
If \(\mathrm{z}+\mathrm{z}^{-1}=1\), then \(\mathrm{z}^{100}+\mathrm{z}^{-100}\) is equal to
1 i
2 \(-\mathrm{i}\)
3 1
4 -1
Explanation:
D Given that, \(\mathrm{z}+\mathrm{z}^{-1}=1\) Then, \(\quad z+\frac{1}{z}=1\) \(\mathrm{z}^2+1-\mathrm{z}=0\) So, \(\quad z=\frac{1 \pm \sqrt{(1)^2-4}}{2}\) \(=\frac{1 \pm \sqrt{3} \mathrm{i}}{2}\) \(\mathrm{z} =\frac{1+\mathrm{i} \sqrt{3}}{2}, \frac{1-\mathrm{i} \sqrt{3}}{2}\) \(\mathrm{z} =-\left(\frac{-1-\mathrm{i} \sqrt{3}}{2}\right),-\left(\frac{-1+\mathrm{i} \sqrt{3}}{2}\right)\) \(z=-\omega^2,-\omega \quad\{\) where \(\omega\) is cube root of unity) Now, \(\quad z^{100}+\frac{1}{z^{100}}=(-\omega)^{100}+\frac{1}{(-\omega)^{100}}\) \(=\omega^{100}+\frac{1}{\omega^{100}}\) \(=\left(\omega^3\right)^{33} \cdot \omega+\) \(=\omega+\frac{1}{\omega}\) \(=\omega+\omega^2\) \(=-1\)
JCECE-2011
Complex Numbers and Quadratic Equation
117847
If \(x=a+b, y=a \omega+b \omega^2\) and \(z=a \omega^2+b \omega\), then \(x y z\) is equal to
1 \((a+b)^3\)
2 \(a^3+b^3\)
3 \(a^3-b^3\)
4 \((a+b)^3-3 a b(a+b)\)
Explanation:
B Given that, \(x=a+b, y=a \omega+b \omega^2 \text { and } z=a \omega^2+b \omega\) \(\text { then, } \quad x \cdot y \cdot z\) \(=(a+b)\left(a \omega+b \omega^2\right)\left(a \omega^2+b \omega\right)\) \(=(a+b)\left(a^2 \omega^3+a b \omega^2+a b \omega^4+b^2 \cdot \omega^3\right)\) \((a+b)\left(a^2+a b \omega^2+a b \omega+b^2\right)\) \((a+b)\left(a^2+b^2+a b\left(\omega^2+\omega\right)\right.\) \((a+b)\left(a^2+b^2-a b\right)\) \(=a^3+b^3\)
117844
If \(\frac{|z-2|}{|z-3|}=2\) represents a circle, then its radius is equal to
1 1
2 \(\frac{1}{3}\)
3 \(\frac{3}{4}\)
4 \(\frac{2}{3}\)
Explanation:
D Given that, \(\frac{|z-2|}{|z-3|}=2\) Let, \(\quad z=x+i y\) Then, \(\quad \frac{|x+i y-2|}{|x+i y-3|}=2\) \(\mid(x-2)+\text { iy }|=2|(x-3)+i y \mid\) \(\sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2}\) Squaring both side - \((x-2)^2+y^2=4\left\{(x-3)^2+y^2\right\}\) \(x^2+4-4 x+y^2=4 x^2+36-24 x+4 y^2\) \(3 x^2+3 y^2-20 x+32=0\) \(x^2+y^2-\frac{20}{3} x+\frac{32}{3}=0\) Adding \(\frac{100}{9}\) both side \(\left(x-\frac{10}{3}\right)^2+y^2+\frac{32}{3}=\frac{100}{9}\) \(\left(x-\frac{10}{3}\right)^2+y^2=\frac{100}{9}-\frac{32}{3}\) \(\left(x-\frac{10}{3}\right)^2+y^2=\frac{4}{9}\)Above equation represent a circle with center \(\left(\frac{10}{3}, 0\right)\) and radius \(\frac{2}{3}\)
JCECE-2014
Complex Numbers and Quadratic Equation
117845
If \(\theta\) is real and \(z_1, z_2\) are connected by \(\mathrm{z}_1^2+\mathrm{z}_2^2+2 \mathrm{z}_1 \mathrm{z}_2 \cos \theta=0\), then triangle with vertices \(0, z_1\) and \(z_2\) is
1 equilateral
2 right angled
3 isosceles
4 None of above
Explanation:
C Given that, \(z_1^2+z_2^2+2 z_1 z_2 \cos \theta=0\) Now, above equation is divided by \(\left(z_2\right)^2\) \(\frac{z_1^2}{z_2^2}+1+\frac{2 z_1}{z_2} \cos \theta=0\) \(\left(\frac{z_1}{z_2}\right)^2+2\left(\frac{z_1}{z_2}\right) \cos \theta+\cos ^2 \theta+\sin ^2 \theta=0\) \(\left(\frac{z_1}{z_2}+\cos \theta\right)^2+\sin ^2 \theta=0\) \(\left.\left(\frac{z_1}{z_2}+\cos \theta\right)^2=-\sin ^2 \theta=1\right)\) \(\frac{z_1}{z_2}+\cos \theta= \pm \sin \theta\) \(\frac{z_1}{z_2}=-\cos \theta \pm \sin \theta\) \(\left|\frac{z_1}{z_2}\right|=\sqrt{(-\cos \theta)^2+\sin ^2 \theta}=\sqrt{1}=1\) \(\left|z_1\right|=\left|z_2\right|\) \(\left|z_1-0\right|=\left|z_2-0\right|\)From above we can say that, triangle with vertices \(0, z_1\) and \(z_2\) is isosceles.
JCECE-2013
Complex Numbers and Quadratic Equation
117846
If \(\mathrm{z}+\mathrm{z}^{-1}=1\), then \(\mathrm{z}^{100}+\mathrm{z}^{-100}\) is equal to
1 i
2 \(-\mathrm{i}\)
3 1
4 -1
Explanation:
D Given that, \(\mathrm{z}+\mathrm{z}^{-1}=1\) Then, \(\quad z+\frac{1}{z}=1\) \(\mathrm{z}^2+1-\mathrm{z}=0\) So, \(\quad z=\frac{1 \pm \sqrt{(1)^2-4}}{2}\) \(=\frac{1 \pm \sqrt{3} \mathrm{i}}{2}\) \(\mathrm{z} =\frac{1+\mathrm{i} \sqrt{3}}{2}, \frac{1-\mathrm{i} \sqrt{3}}{2}\) \(\mathrm{z} =-\left(\frac{-1-\mathrm{i} \sqrt{3}}{2}\right),-\left(\frac{-1+\mathrm{i} \sqrt{3}}{2}\right)\) \(z=-\omega^2,-\omega \quad\{\) where \(\omega\) is cube root of unity) Now, \(\quad z^{100}+\frac{1}{z^{100}}=(-\omega)^{100}+\frac{1}{(-\omega)^{100}}\) \(=\omega^{100}+\frac{1}{\omega^{100}}\) \(=\left(\omega^3\right)^{33} \cdot \omega+\) \(=\omega+\frac{1}{\omega}\) \(=\omega+\omega^2\) \(=-1\)
JCECE-2011
Complex Numbers and Quadratic Equation
117847
If \(x=a+b, y=a \omega+b \omega^2\) and \(z=a \omega^2+b \omega\), then \(x y z\) is equal to
1 \((a+b)^3\)
2 \(a^3+b^3\)
3 \(a^3-b^3\)
4 \((a+b)^3-3 a b(a+b)\)
Explanation:
B Given that, \(x=a+b, y=a \omega+b \omega^2 \text { and } z=a \omega^2+b \omega\) \(\text { then, } \quad x \cdot y \cdot z\) \(=(a+b)\left(a \omega+b \omega^2\right)\left(a \omega^2+b \omega\right)\) \(=(a+b)\left(a^2 \omega^3+a b \omega^2+a b \omega^4+b^2 \cdot \omega^3\right)\) \((a+b)\left(a^2+a b \omega^2+a b \omega+b^2\right)\) \((a+b)\left(a^2+b^2+a b\left(\omega^2+\omega\right)\right.\) \((a+b)\left(a^2+b^2-a b\right)\) \(=a^3+b^3\)
117844
If \(\frac{|z-2|}{|z-3|}=2\) represents a circle, then its radius is equal to
1 1
2 \(\frac{1}{3}\)
3 \(\frac{3}{4}\)
4 \(\frac{2}{3}\)
Explanation:
D Given that, \(\frac{|z-2|}{|z-3|}=2\) Let, \(\quad z=x+i y\) Then, \(\quad \frac{|x+i y-2|}{|x+i y-3|}=2\) \(\mid(x-2)+\text { iy }|=2|(x-3)+i y \mid\) \(\sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2}\) Squaring both side - \((x-2)^2+y^2=4\left\{(x-3)^2+y^2\right\}\) \(x^2+4-4 x+y^2=4 x^2+36-24 x+4 y^2\) \(3 x^2+3 y^2-20 x+32=0\) \(x^2+y^2-\frac{20}{3} x+\frac{32}{3}=0\) Adding \(\frac{100}{9}\) both side \(\left(x-\frac{10}{3}\right)^2+y^2+\frac{32}{3}=\frac{100}{9}\) \(\left(x-\frac{10}{3}\right)^2+y^2=\frac{100}{9}-\frac{32}{3}\) \(\left(x-\frac{10}{3}\right)^2+y^2=\frac{4}{9}\)Above equation represent a circle with center \(\left(\frac{10}{3}, 0\right)\) and radius \(\frac{2}{3}\)
JCECE-2014
Complex Numbers and Quadratic Equation
117845
If \(\theta\) is real and \(z_1, z_2\) are connected by \(\mathrm{z}_1^2+\mathrm{z}_2^2+2 \mathrm{z}_1 \mathrm{z}_2 \cos \theta=0\), then triangle with vertices \(0, z_1\) and \(z_2\) is
1 equilateral
2 right angled
3 isosceles
4 None of above
Explanation:
C Given that, \(z_1^2+z_2^2+2 z_1 z_2 \cos \theta=0\) Now, above equation is divided by \(\left(z_2\right)^2\) \(\frac{z_1^2}{z_2^2}+1+\frac{2 z_1}{z_2} \cos \theta=0\) \(\left(\frac{z_1}{z_2}\right)^2+2\left(\frac{z_1}{z_2}\right) \cos \theta+\cos ^2 \theta+\sin ^2 \theta=0\) \(\left(\frac{z_1}{z_2}+\cos \theta\right)^2+\sin ^2 \theta=0\) \(\left.\left(\frac{z_1}{z_2}+\cos \theta\right)^2=-\sin ^2 \theta=1\right)\) \(\frac{z_1}{z_2}+\cos \theta= \pm \sin \theta\) \(\frac{z_1}{z_2}=-\cos \theta \pm \sin \theta\) \(\left|\frac{z_1}{z_2}\right|=\sqrt{(-\cos \theta)^2+\sin ^2 \theta}=\sqrt{1}=1\) \(\left|z_1\right|=\left|z_2\right|\) \(\left|z_1-0\right|=\left|z_2-0\right|\)From above we can say that, triangle with vertices \(0, z_1\) and \(z_2\) is isosceles.
JCECE-2013
Complex Numbers and Quadratic Equation
117846
If \(\mathrm{z}+\mathrm{z}^{-1}=1\), then \(\mathrm{z}^{100}+\mathrm{z}^{-100}\) is equal to
1 i
2 \(-\mathrm{i}\)
3 1
4 -1
Explanation:
D Given that, \(\mathrm{z}+\mathrm{z}^{-1}=1\) Then, \(\quad z+\frac{1}{z}=1\) \(\mathrm{z}^2+1-\mathrm{z}=0\) So, \(\quad z=\frac{1 \pm \sqrt{(1)^2-4}}{2}\) \(=\frac{1 \pm \sqrt{3} \mathrm{i}}{2}\) \(\mathrm{z} =\frac{1+\mathrm{i} \sqrt{3}}{2}, \frac{1-\mathrm{i} \sqrt{3}}{2}\) \(\mathrm{z} =-\left(\frac{-1-\mathrm{i} \sqrt{3}}{2}\right),-\left(\frac{-1+\mathrm{i} \sqrt{3}}{2}\right)\) \(z=-\omega^2,-\omega \quad\{\) where \(\omega\) is cube root of unity) Now, \(\quad z^{100}+\frac{1}{z^{100}}=(-\omega)^{100}+\frac{1}{(-\omega)^{100}}\) \(=\omega^{100}+\frac{1}{\omega^{100}}\) \(=\left(\omega^3\right)^{33} \cdot \omega+\) \(=\omega+\frac{1}{\omega}\) \(=\omega+\omega^2\) \(=-1\)
JCECE-2011
Complex Numbers and Quadratic Equation
117847
If \(x=a+b, y=a \omega+b \omega^2\) and \(z=a \omega^2+b \omega\), then \(x y z\) is equal to
1 \((a+b)^3\)
2 \(a^3+b^3\)
3 \(a^3-b^3\)
4 \((a+b)^3-3 a b(a+b)\)
Explanation:
B Given that, \(x=a+b, y=a \omega+b \omega^2 \text { and } z=a \omega^2+b \omega\) \(\text { then, } \quad x \cdot y \cdot z\) \(=(a+b)\left(a \omega+b \omega^2\right)\left(a \omega^2+b \omega\right)\) \(=(a+b)\left(a^2 \omega^3+a b \omega^2+a b \omega^4+b^2 \cdot \omega^3\right)\) \((a+b)\left(a^2+a b \omega^2+a b \omega+b^2\right)\) \((a+b)\left(a^2+b^2+a b\left(\omega^2+\omega\right)\right.\) \((a+b)\left(a^2+b^2-a b\right)\) \(=a^3+b^3\)
