117835
A complex number \(\mathrm{z}\) is such that arg \(\left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}\). The points representing this complex number will lie on
1 an ellipse
2 a parabola
3 a circle
4 a straight line
Explanation:
C Given, \(\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}\) Let \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(\arg \left(\frac{\mathrm{x}+\mathrm{iy}-2}{\mathrm{x}+\mathrm{iy}+2}\right)=\frac{\pi}{3}\) \(\arg \left(\frac{\mathrm{x}-2+\mathrm{iy}}{\mathrm{x}+2+\mathrm{iy}}\right)=\frac{\pi}{3}\) \(\arg (x-2+i y)-\arg (x+2+i y)=\frac{\pi}{3}\) \(\tan ^{-1}\left(\frac{\left(\frac{y}{x-2}\right)-\left(\frac{y}{x+2}\right)}{1+\frac{y}{(x-2)} \times \frac{y}{(x+2)}}\right)=\frac{\pi}{3}\left\{\begin{array}{l}\because \tan ^{-1} x-\tan ^{-1} y \\ =\tan ^{-1}\left(\frac{x-y}{1-x y}\right)\end{array}\right\}\) \(=\tan ^{-1}\left(\frac{x-y}{1-x y}\right)\) \(\frac{4 \mathrm{y}}{\mathrm{x}^2+\mathrm{y}^2-4}=\tan \frac{\pi}{3}\) \(\frac{4 \mathrm{y}}{\mathrm{x}^2+\mathrm{y}^2-4}=\sqrt{3}\) \(4 \mathrm{y}=\sqrt{3}\left(\mathrm{x}^2+\mathrm{y}^2-4\right)\) \(4 y=\sqrt{3} x^2+\sqrt{3} y^2-4 \sqrt{3}\) \(\sqrt{3} x^2+\sqrt{3} y^2-4 y-4 \sqrt{3}=0\) \(\sqrt{3}\left(x^2+y^2\right)-4 y-4 \sqrt{3}=0\)Which is an equation of a circle.
VITEEE-2013
Complex Numbers and Quadratic Equation
117836
If \(\omega=\frac{-1+\sqrt{3} i}{2}\) then \(\left(3+\omega+3 \omega^2\right)^4\) is
1 16
2 -16
3 \(16 \omega\)
4 \(16 \omega^2\)
Explanation:
C Given, \(\omega=\frac{-1+\sqrt{3} \mathrm{i}}{2}\) We know that \(1+\omega+\omega^2=0 \text { and } \omega^3=1\) Then \(\left(3+3 \omega+3 \omega^2\right)^4\) \(=\left(3+\omega+3 \omega^2+2 \omega-2 \omega\right)^4\) \(=\left(3+3 \omega+3 \omega^2-2 \omega\right)^4\) \(=\left[3\left(1+\omega+\omega^2\right)-2 \omega\right]^4\) \(=(0-2 \omega)^4\) \(=16 \omega^4\) \(=16 \omega^3 \cdot \omega \quad \because\left(\omega^3=1\right)\) \(=16 \omega\)
VITEEE-2019
Complex Numbers and Quadratic Equation
117837
If \(x+i y=(1-i \sqrt{3})^{100}\), then find \((x, y)\).
1 \(\left(2^{99}, 2^{99} \sqrt{3}\right)\)
2 \(\left(2^{99},-2^{99} \sqrt{3}\right)\)
3 \(\left(-2^{99}, 2^{99} \sqrt{3}\right)\)
4 None of these
Explanation:
C Given, \(\text { If } x+i y =(1-i \sqrt{3})^{100},\) \((1-i \sqrt{3})^{100} =2^{100}\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{100}\) \(= 2^{100} \omega^{100}=2^{100} \omega \quad\left(\because \omega=-\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)\) \(= 2^{100}\left(-\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)=-2^{99}+2^{99} \sqrt{3} \mathrm{i}\) Now, \(x+i y=(1-i \sqrt{3})^{100}\) \(=-2^{99}+2^{99} \sqrt{3} \mathrm{i}\) Comparing real and imaginary parts- \(\Rightarrow \mathrm{x}=-2^{99}, \mathrm{y}=2^{99} \sqrt{3} \mathrm{i}\) \(\therefore(\mathrm{x}, \mathrm{y})=\left(-2^{99}, 2^{99} \sqrt{3}\right)\)
VITEEE-2011
Complex Numbers and Quadratic Equation
117838
If \(z=\frac{1-i \sqrt{3}}{1+i \sqrt{3}}\), then \(\arg (z)\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
117835
A complex number \(\mathrm{z}\) is such that arg \(\left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}\). The points representing this complex number will lie on
1 an ellipse
2 a parabola
3 a circle
4 a straight line
Explanation:
C Given, \(\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}\) Let \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(\arg \left(\frac{\mathrm{x}+\mathrm{iy}-2}{\mathrm{x}+\mathrm{iy}+2}\right)=\frac{\pi}{3}\) \(\arg \left(\frac{\mathrm{x}-2+\mathrm{iy}}{\mathrm{x}+2+\mathrm{iy}}\right)=\frac{\pi}{3}\) \(\arg (x-2+i y)-\arg (x+2+i y)=\frac{\pi}{3}\) \(\tan ^{-1}\left(\frac{\left(\frac{y}{x-2}\right)-\left(\frac{y}{x+2}\right)}{1+\frac{y}{(x-2)} \times \frac{y}{(x+2)}}\right)=\frac{\pi}{3}\left\{\begin{array}{l}\because \tan ^{-1} x-\tan ^{-1} y \\ =\tan ^{-1}\left(\frac{x-y}{1-x y}\right)\end{array}\right\}\) \(=\tan ^{-1}\left(\frac{x-y}{1-x y}\right)\) \(\frac{4 \mathrm{y}}{\mathrm{x}^2+\mathrm{y}^2-4}=\tan \frac{\pi}{3}\) \(\frac{4 \mathrm{y}}{\mathrm{x}^2+\mathrm{y}^2-4}=\sqrt{3}\) \(4 \mathrm{y}=\sqrt{3}\left(\mathrm{x}^2+\mathrm{y}^2-4\right)\) \(4 y=\sqrt{3} x^2+\sqrt{3} y^2-4 \sqrt{3}\) \(\sqrt{3} x^2+\sqrt{3} y^2-4 y-4 \sqrt{3}=0\) \(\sqrt{3}\left(x^2+y^2\right)-4 y-4 \sqrt{3}=0\)Which is an equation of a circle.
VITEEE-2013
Complex Numbers and Quadratic Equation
117836
If \(\omega=\frac{-1+\sqrt{3} i}{2}\) then \(\left(3+\omega+3 \omega^2\right)^4\) is
1 16
2 -16
3 \(16 \omega\)
4 \(16 \omega^2\)
Explanation:
C Given, \(\omega=\frac{-1+\sqrt{3} \mathrm{i}}{2}\) We know that \(1+\omega+\omega^2=0 \text { and } \omega^3=1\) Then \(\left(3+3 \omega+3 \omega^2\right)^4\) \(=\left(3+\omega+3 \omega^2+2 \omega-2 \omega\right)^4\) \(=\left(3+3 \omega+3 \omega^2-2 \omega\right)^4\) \(=\left[3\left(1+\omega+\omega^2\right)-2 \omega\right]^4\) \(=(0-2 \omega)^4\) \(=16 \omega^4\) \(=16 \omega^3 \cdot \omega \quad \because\left(\omega^3=1\right)\) \(=16 \omega\)
VITEEE-2019
Complex Numbers and Quadratic Equation
117837
If \(x+i y=(1-i \sqrt{3})^{100}\), then find \((x, y)\).
1 \(\left(2^{99}, 2^{99} \sqrt{3}\right)\)
2 \(\left(2^{99},-2^{99} \sqrt{3}\right)\)
3 \(\left(-2^{99}, 2^{99} \sqrt{3}\right)\)
4 None of these
Explanation:
C Given, \(\text { If } x+i y =(1-i \sqrt{3})^{100},\) \((1-i \sqrt{3})^{100} =2^{100}\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{100}\) \(= 2^{100} \omega^{100}=2^{100} \omega \quad\left(\because \omega=-\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)\) \(= 2^{100}\left(-\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)=-2^{99}+2^{99} \sqrt{3} \mathrm{i}\) Now, \(x+i y=(1-i \sqrt{3})^{100}\) \(=-2^{99}+2^{99} \sqrt{3} \mathrm{i}\) Comparing real and imaginary parts- \(\Rightarrow \mathrm{x}=-2^{99}, \mathrm{y}=2^{99} \sqrt{3} \mathrm{i}\) \(\therefore(\mathrm{x}, \mathrm{y})=\left(-2^{99}, 2^{99} \sqrt{3}\right)\)
VITEEE-2011
Complex Numbers and Quadratic Equation
117838
If \(z=\frac{1-i \sqrt{3}}{1+i \sqrt{3}}\), then \(\arg (z)\) is
117835
A complex number \(\mathrm{z}\) is such that arg \(\left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}\). The points representing this complex number will lie on
1 an ellipse
2 a parabola
3 a circle
4 a straight line
Explanation:
C Given, \(\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}\) Let \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(\arg \left(\frac{\mathrm{x}+\mathrm{iy}-2}{\mathrm{x}+\mathrm{iy}+2}\right)=\frac{\pi}{3}\) \(\arg \left(\frac{\mathrm{x}-2+\mathrm{iy}}{\mathrm{x}+2+\mathrm{iy}}\right)=\frac{\pi}{3}\) \(\arg (x-2+i y)-\arg (x+2+i y)=\frac{\pi}{3}\) \(\tan ^{-1}\left(\frac{\left(\frac{y}{x-2}\right)-\left(\frac{y}{x+2}\right)}{1+\frac{y}{(x-2)} \times \frac{y}{(x+2)}}\right)=\frac{\pi}{3}\left\{\begin{array}{l}\because \tan ^{-1} x-\tan ^{-1} y \\ =\tan ^{-1}\left(\frac{x-y}{1-x y}\right)\end{array}\right\}\) \(=\tan ^{-1}\left(\frac{x-y}{1-x y}\right)\) \(\frac{4 \mathrm{y}}{\mathrm{x}^2+\mathrm{y}^2-4}=\tan \frac{\pi}{3}\) \(\frac{4 \mathrm{y}}{\mathrm{x}^2+\mathrm{y}^2-4}=\sqrt{3}\) \(4 \mathrm{y}=\sqrt{3}\left(\mathrm{x}^2+\mathrm{y}^2-4\right)\) \(4 y=\sqrt{3} x^2+\sqrt{3} y^2-4 \sqrt{3}\) \(\sqrt{3} x^2+\sqrt{3} y^2-4 y-4 \sqrt{3}=0\) \(\sqrt{3}\left(x^2+y^2\right)-4 y-4 \sqrt{3}=0\)Which is an equation of a circle.
VITEEE-2013
Complex Numbers and Quadratic Equation
117836
If \(\omega=\frac{-1+\sqrt{3} i}{2}\) then \(\left(3+\omega+3 \omega^2\right)^4\) is
1 16
2 -16
3 \(16 \omega\)
4 \(16 \omega^2\)
Explanation:
C Given, \(\omega=\frac{-1+\sqrt{3} \mathrm{i}}{2}\) We know that \(1+\omega+\omega^2=0 \text { and } \omega^3=1\) Then \(\left(3+3 \omega+3 \omega^2\right)^4\) \(=\left(3+\omega+3 \omega^2+2 \omega-2 \omega\right)^4\) \(=\left(3+3 \omega+3 \omega^2-2 \omega\right)^4\) \(=\left[3\left(1+\omega+\omega^2\right)-2 \omega\right]^4\) \(=(0-2 \omega)^4\) \(=16 \omega^4\) \(=16 \omega^3 \cdot \omega \quad \because\left(\omega^3=1\right)\) \(=16 \omega\)
VITEEE-2019
Complex Numbers and Quadratic Equation
117837
If \(x+i y=(1-i \sqrt{3})^{100}\), then find \((x, y)\).
1 \(\left(2^{99}, 2^{99} \sqrt{3}\right)\)
2 \(\left(2^{99},-2^{99} \sqrt{3}\right)\)
3 \(\left(-2^{99}, 2^{99} \sqrt{3}\right)\)
4 None of these
Explanation:
C Given, \(\text { If } x+i y =(1-i \sqrt{3})^{100},\) \((1-i \sqrt{3})^{100} =2^{100}\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{100}\) \(= 2^{100} \omega^{100}=2^{100} \omega \quad\left(\because \omega=-\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)\) \(= 2^{100}\left(-\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)=-2^{99}+2^{99} \sqrt{3} \mathrm{i}\) Now, \(x+i y=(1-i \sqrt{3})^{100}\) \(=-2^{99}+2^{99} \sqrt{3} \mathrm{i}\) Comparing real and imaginary parts- \(\Rightarrow \mathrm{x}=-2^{99}, \mathrm{y}=2^{99} \sqrt{3} \mathrm{i}\) \(\therefore(\mathrm{x}, \mathrm{y})=\left(-2^{99}, 2^{99} \sqrt{3}\right)\)
VITEEE-2011
Complex Numbers and Quadratic Equation
117838
If \(z=\frac{1-i \sqrt{3}}{1+i \sqrt{3}}\), then \(\arg (z)\) is
117835
A complex number \(\mathrm{z}\) is such that arg \(\left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}\). The points representing this complex number will lie on
1 an ellipse
2 a parabola
3 a circle
4 a straight line
Explanation:
C Given, \(\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}\) Let \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(\arg \left(\frac{\mathrm{x}+\mathrm{iy}-2}{\mathrm{x}+\mathrm{iy}+2}\right)=\frac{\pi}{3}\) \(\arg \left(\frac{\mathrm{x}-2+\mathrm{iy}}{\mathrm{x}+2+\mathrm{iy}}\right)=\frac{\pi}{3}\) \(\arg (x-2+i y)-\arg (x+2+i y)=\frac{\pi}{3}\) \(\tan ^{-1}\left(\frac{\left(\frac{y}{x-2}\right)-\left(\frac{y}{x+2}\right)}{1+\frac{y}{(x-2)} \times \frac{y}{(x+2)}}\right)=\frac{\pi}{3}\left\{\begin{array}{l}\because \tan ^{-1} x-\tan ^{-1} y \\ =\tan ^{-1}\left(\frac{x-y}{1-x y}\right)\end{array}\right\}\) \(=\tan ^{-1}\left(\frac{x-y}{1-x y}\right)\) \(\frac{4 \mathrm{y}}{\mathrm{x}^2+\mathrm{y}^2-4}=\tan \frac{\pi}{3}\) \(\frac{4 \mathrm{y}}{\mathrm{x}^2+\mathrm{y}^2-4}=\sqrt{3}\) \(4 \mathrm{y}=\sqrt{3}\left(\mathrm{x}^2+\mathrm{y}^2-4\right)\) \(4 y=\sqrt{3} x^2+\sqrt{3} y^2-4 \sqrt{3}\) \(\sqrt{3} x^2+\sqrt{3} y^2-4 y-4 \sqrt{3}=0\) \(\sqrt{3}\left(x^2+y^2\right)-4 y-4 \sqrt{3}=0\)Which is an equation of a circle.
VITEEE-2013
Complex Numbers and Quadratic Equation
117836
If \(\omega=\frac{-1+\sqrt{3} i}{2}\) then \(\left(3+\omega+3 \omega^2\right)^4\) is
1 16
2 -16
3 \(16 \omega\)
4 \(16 \omega^2\)
Explanation:
C Given, \(\omega=\frac{-1+\sqrt{3} \mathrm{i}}{2}\) We know that \(1+\omega+\omega^2=0 \text { and } \omega^3=1\) Then \(\left(3+3 \omega+3 \omega^2\right)^4\) \(=\left(3+\omega+3 \omega^2+2 \omega-2 \omega\right)^4\) \(=\left(3+3 \omega+3 \omega^2-2 \omega\right)^4\) \(=\left[3\left(1+\omega+\omega^2\right)-2 \omega\right]^4\) \(=(0-2 \omega)^4\) \(=16 \omega^4\) \(=16 \omega^3 \cdot \omega \quad \because\left(\omega^3=1\right)\) \(=16 \omega\)
VITEEE-2019
Complex Numbers and Quadratic Equation
117837
If \(x+i y=(1-i \sqrt{3})^{100}\), then find \((x, y)\).
1 \(\left(2^{99}, 2^{99} \sqrt{3}\right)\)
2 \(\left(2^{99},-2^{99} \sqrt{3}\right)\)
3 \(\left(-2^{99}, 2^{99} \sqrt{3}\right)\)
4 None of these
Explanation:
C Given, \(\text { If } x+i y =(1-i \sqrt{3})^{100},\) \((1-i \sqrt{3})^{100} =2^{100}\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{100}\) \(= 2^{100} \omega^{100}=2^{100} \omega \quad\left(\because \omega=-\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)\) \(= 2^{100}\left(-\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)=-2^{99}+2^{99} \sqrt{3} \mathrm{i}\) Now, \(x+i y=(1-i \sqrt{3})^{100}\) \(=-2^{99}+2^{99} \sqrt{3} \mathrm{i}\) Comparing real and imaginary parts- \(\Rightarrow \mathrm{x}=-2^{99}, \mathrm{y}=2^{99} \sqrt{3} \mathrm{i}\) \(\therefore(\mathrm{x}, \mathrm{y})=\left(-2^{99}, 2^{99} \sqrt{3}\right)\)
VITEEE-2011
Complex Numbers and Quadratic Equation
117838
If \(z=\frac{1-i \sqrt{3}}{1+i \sqrt{3}}\), then \(\arg (z)\) is