117963
If \(P(x, y)\) denotes \(z=x+\) iy in Argand's plane and \(\left|\frac{z-1}{z+2 i}\right|=1\), then the locus of \(P\) is a/an
1 hyperbola
2 ellipse
3 circle
4 straight line
Explanation:
D Given, \(z=x+\) iy and \(\left|\frac{z-1}{z+2 i}\right|=1\) \(\text { Then, }\left|\frac{(x+i y)-1}{(x+i y)+2 i}\right|=1\) \(|(x-1)+i y|=|x+i(y+2)|\) \(\sqrt{(x-1)^2+y^2}=\sqrt{x^2+(y+2)^2}\) Squaring both side \((x-1)^2+y^2=x^2+(y+2)^2\) \(x^2+1-2 x+y^2=x^2+y^2+2 y+4\) \(2 x+2 y+3=0\)Above equation shows a straight line so, locus of \(\mathrm{P}\) is a straight line.
Karnataka CET-2011
Complex Numbers and Quadratic Equation
117964
In Argand's plane, the point corresponding to \(\frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}\) lies in
1 quadrant I
2 quadrant II
3 quadrant III
4 quadrant IV
Explanation:
D \( \frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}\) \(=\frac{1+i-i \sqrt{3}+\sqrt{3}}{(\sqrt{3}+i)}\) \(=\frac{(i+\sqrt{3})+(1-i \sqrt{3})}{(\sqrt{3}+i)}\) \(=1+\frac{(1-i \sqrt{3})}{(\sqrt{3}+i)}\) \(=1+\frac{(1-i \sqrt{3})}{(\sqrt{3}+i)} \times \frac{(\sqrt{3}-i)}{(\sqrt{3}-i)}\) \(=1+\frac{\sqrt{3}-3 i-i-\sqrt{3}}{3+1}\) \(=1-\frac{4 i}{4}=1-i\) \(\text { So, the point lies in quadrant IV. }\)So, the point lies in quadrant IV.
Karnataka CET-2011
Complex Numbers and Quadratic Equation
117965
The complex number \(\frac{(-\sqrt{3}+3 i)(1-i)}{(3+\sqrt{3} i) i(\sqrt{3}+\sqrt{3} i)}\) when represented in the Argand diagram is
1 in the second quadrant
2 in the first quadrant
3 on the \(\mathrm{Y}\)-axis (imaginary axis)
4 on the \(\mathrm{X}\)-axis (real axis)
Explanation:
C Given complex number is- \(\frac{(-\sqrt{3}+3 i)(1-i)}{(3+\sqrt{3} i) i(\sqrt{3}+\sqrt{3} i)}\) Let \(z=\frac{(-\sqrt{3}+3 i)(1-i)}{(3+\sqrt{3} i) i(\sqrt{3}+\sqrt{3} i)}\) \(=\frac{\sqrt{3}(-1+\sqrt{3} i)(1-i)}{(\sqrt{3})^2(\sqrt{3}+i) i(1+i)}\) \(=\frac{1}{\sqrt{3}} \frac{(-1+\sqrt{3} i)(1-i)}{(-1+\sqrt{3} i)(1+i)}\) \(z=\frac{1}{\sqrt{3}} \frac{(1-i)}{(1+i)}\) Rationalize the above complex number \(z=\frac{1}{\sqrt{3}} \frac{(1-i)(1-i)}{(1+i)(1-i)}=\frac{1}{\sqrt{3}} \frac{(1-1-2 i)}{1+1}\) \(z=\frac{-1}{\sqrt{3}} i\)From above it is clear given complex number lies on \(\mathrm{Y}\)-axis.
117963
If \(P(x, y)\) denotes \(z=x+\) iy in Argand's plane and \(\left|\frac{z-1}{z+2 i}\right|=1\), then the locus of \(P\) is a/an
1 hyperbola
2 ellipse
3 circle
4 straight line
Explanation:
D Given, \(z=x+\) iy and \(\left|\frac{z-1}{z+2 i}\right|=1\) \(\text { Then, }\left|\frac{(x+i y)-1}{(x+i y)+2 i}\right|=1\) \(|(x-1)+i y|=|x+i(y+2)|\) \(\sqrt{(x-1)^2+y^2}=\sqrt{x^2+(y+2)^2}\) Squaring both side \((x-1)^2+y^2=x^2+(y+2)^2\) \(x^2+1-2 x+y^2=x^2+y^2+2 y+4\) \(2 x+2 y+3=0\)Above equation shows a straight line so, locus of \(\mathrm{P}\) is a straight line.
Karnataka CET-2011
Complex Numbers and Quadratic Equation
117964
In Argand's plane, the point corresponding to \(\frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}\) lies in
1 quadrant I
2 quadrant II
3 quadrant III
4 quadrant IV
Explanation:
D \( \frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}\) \(=\frac{1+i-i \sqrt{3}+\sqrt{3}}{(\sqrt{3}+i)}\) \(=\frac{(i+\sqrt{3})+(1-i \sqrt{3})}{(\sqrt{3}+i)}\) \(=1+\frac{(1-i \sqrt{3})}{(\sqrt{3}+i)}\) \(=1+\frac{(1-i \sqrt{3})}{(\sqrt{3}+i)} \times \frac{(\sqrt{3}-i)}{(\sqrt{3}-i)}\) \(=1+\frac{\sqrt{3}-3 i-i-\sqrt{3}}{3+1}\) \(=1-\frac{4 i}{4}=1-i\) \(\text { So, the point lies in quadrant IV. }\)So, the point lies in quadrant IV.
Karnataka CET-2011
Complex Numbers and Quadratic Equation
117965
The complex number \(\frac{(-\sqrt{3}+3 i)(1-i)}{(3+\sqrt{3} i) i(\sqrt{3}+\sqrt{3} i)}\) when represented in the Argand diagram is
1 in the second quadrant
2 in the first quadrant
3 on the \(\mathrm{Y}\)-axis (imaginary axis)
4 on the \(\mathrm{X}\)-axis (real axis)
Explanation:
C Given complex number is- \(\frac{(-\sqrt{3}+3 i)(1-i)}{(3+\sqrt{3} i) i(\sqrt{3}+\sqrt{3} i)}\) Let \(z=\frac{(-\sqrt{3}+3 i)(1-i)}{(3+\sqrt{3} i) i(\sqrt{3}+\sqrt{3} i)}\) \(=\frac{\sqrt{3}(-1+\sqrt{3} i)(1-i)}{(\sqrt{3})^2(\sqrt{3}+i) i(1+i)}\) \(=\frac{1}{\sqrt{3}} \frac{(-1+\sqrt{3} i)(1-i)}{(-1+\sqrt{3} i)(1+i)}\) \(z=\frac{1}{\sqrt{3}} \frac{(1-i)}{(1+i)}\) Rationalize the above complex number \(z=\frac{1}{\sqrt{3}} \frac{(1-i)(1-i)}{(1+i)(1-i)}=\frac{1}{\sqrt{3}} \frac{(1-1-2 i)}{1+1}\) \(z=\frac{-1}{\sqrt{3}} i\)From above it is clear given complex number lies on \(\mathrm{Y}\)-axis.
117963
If \(P(x, y)\) denotes \(z=x+\) iy in Argand's plane and \(\left|\frac{z-1}{z+2 i}\right|=1\), then the locus of \(P\) is a/an
1 hyperbola
2 ellipse
3 circle
4 straight line
Explanation:
D Given, \(z=x+\) iy and \(\left|\frac{z-1}{z+2 i}\right|=1\) \(\text { Then, }\left|\frac{(x+i y)-1}{(x+i y)+2 i}\right|=1\) \(|(x-1)+i y|=|x+i(y+2)|\) \(\sqrt{(x-1)^2+y^2}=\sqrt{x^2+(y+2)^2}\) Squaring both side \((x-1)^2+y^2=x^2+(y+2)^2\) \(x^2+1-2 x+y^2=x^2+y^2+2 y+4\) \(2 x+2 y+3=0\)Above equation shows a straight line so, locus of \(\mathrm{P}\) is a straight line.
Karnataka CET-2011
Complex Numbers and Quadratic Equation
117964
In Argand's plane, the point corresponding to \(\frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}\) lies in
1 quadrant I
2 quadrant II
3 quadrant III
4 quadrant IV
Explanation:
D \( \frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}\) \(=\frac{1+i-i \sqrt{3}+\sqrt{3}}{(\sqrt{3}+i)}\) \(=\frac{(i+\sqrt{3})+(1-i \sqrt{3})}{(\sqrt{3}+i)}\) \(=1+\frac{(1-i \sqrt{3})}{(\sqrt{3}+i)}\) \(=1+\frac{(1-i \sqrt{3})}{(\sqrt{3}+i)} \times \frac{(\sqrt{3}-i)}{(\sqrt{3}-i)}\) \(=1+\frac{\sqrt{3}-3 i-i-\sqrt{3}}{3+1}\) \(=1-\frac{4 i}{4}=1-i\) \(\text { So, the point lies in quadrant IV. }\)So, the point lies in quadrant IV.
Karnataka CET-2011
Complex Numbers and Quadratic Equation
117965
The complex number \(\frac{(-\sqrt{3}+3 i)(1-i)}{(3+\sqrt{3} i) i(\sqrt{3}+\sqrt{3} i)}\) when represented in the Argand diagram is
1 in the second quadrant
2 in the first quadrant
3 on the \(\mathrm{Y}\)-axis (imaginary axis)
4 on the \(\mathrm{X}\)-axis (real axis)
Explanation:
C Given complex number is- \(\frac{(-\sqrt{3}+3 i)(1-i)}{(3+\sqrt{3} i) i(\sqrt{3}+\sqrt{3} i)}\) Let \(z=\frac{(-\sqrt{3}+3 i)(1-i)}{(3+\sqrt{3} i) i(\sqrt{3}+\sqrt{3} i)}\) \(=\frac{\sqrt{3}(-1+\sqrt{3} i)(1-i)}{(\sqrt{3})^2(\sqrt{3}+i) i(1+i)}\) \(=\frac{1}{\sqrt{3}} \frac{(-1+\sqrt{3} i)(1-i)}{(-1+\sqrt{3} i)(1+i)}\) \(z=\frac{1}{\sqrt{3}} \frac{(1-i)}{(1+i)}\) Rationalize the above complex number \(z=\frac{1}{\sqrt{3}} \frac{(1-i)(1-i)}{(1+i)(1-i)}=\frac{1}{\sqrt{3}} \frac{(1-1-2 i)}{1+1}\) \(z=\frac{-1}{\sqrt{3}} i\)From above it is clear given complex number lies on \(\mathrm{Y}\)-axis.
117963
If \(P(x, y)\) denotes \(z=x+\) iy in Argand's plane and \(\left|\frac{z-1}{z+2 i}\right|=1\), then the locus of \(P\) is a/an
1 hyperbola
2 ellipse
3 circle
4 straight line
Explanation:
D Given, \(z=x+\) iy and \(\left|\frac{z-1}{z+2 i}\right|=1\) \(\text { Then, }\left|\frac{(x+i y)-1}{(x+i y)+2 i}\right|=1\) \(|(x-1)+i y|=|x+i(y+2)|\) \(\sqrt{(x-1)^2+y^2}=\sqrt{x^2+(y+2)^2}\) Squaring both side \((x-1)^2+y^2=x^2+(y+2)^2\) \(x^2+1-2 x+y^2=x^2+y^2+2 y+4\) \(2 x+2 y+3=0\)Above equation shows a straight line so, locus of \(\mathrm{P}\) is a straight line.
Karnataka CET-2011
Complex Numbers and Quadratic Equation
117964
In Argand's plane, the point corresponding to \(\frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}\) lies in
1 quadrant I
2 quadrant II
3 quadrant III
4 quadrant IV
Explanation:
D \( \frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}\) \(=\frac{1+i-i \sqrt{3}+\sqrt{3}}{(\sqrt{3}+i)}\) \(=\frac{(i+\sqrt{3})+(1-i \sqrt{3})}{(\sqrt{3}+i)}\) \(=1+\frac{(1-i \sqrt{3})}{(\sqrt{3}+i)}\) \(=1+\frac{(1-i \sqrt{3})}{(\sqrt{3}+i)} \times \frac{(\sqrt{3}-i)}{(\sqrt{3}-i)}\) \(=1+\frac{\sqrt{3}-3 i-i-\sqrt{3}}{3+1}\) \(=1-\frac{4 i}{4}=1-i\) \(\text { So, the point lies in quadrant IV. }\)So, the point lies in quadrant IV.
Karnataka CET-2011
Complex Numbers and Quadratic Equation
117965
The complex number \(\frac{(-\sqrt{3}+3 i)(1-i)}{(3+\sqrt{3} i) i(\sqrt{3}+\sqrt{3} i)}\) when represented in the Argand diagram is
1 in the second quadrant
2 in the first quadrant
3 on the \(\mathrm{Y}\)-axis (imaginary axis)
4 on the \(\mathrm{X}\)-axis (real axis)
Explanation:
C Given complex number is- \(\frac{(-\sqrt{3}+3 i)(1-i)}{(3+\sqrt{3} i) i(\sqrt{3}+\sqrt{3} i)}\) Let \(z=\frac{(-\sqrt{3}+3 i)(1-i)}{(3+\sqrt{3} i) i(\sqrt{3}+\sqrt{3} i)}\) \(=\frac{\sqrt{3}(-1+\sqrt{3} i)(1-i)}{(\sqrt{3})^2(\sqrt{3}+i) i(1+i)}\) \(=\frac{1}{\sqrt{3}} \frac{(-1+\sqrt{3} i)(1-i)}{(-1+\sqrt{3} i)(1+i)}\) \(z=\frac{1}{\sqrt{3}} \frac{(1-i)}{(1+i)}\) Rationalize the above complex number \(z=\frac{1}{\sqrt{3}} \frac{(1-i)(1-i)}{(1+i)(1-i)}=\frac{1}{\sqrt{3}} \frac{(1-1-2 i)}{1+1}\) \(z=\frac{-1}{\sqrt{3}} i\)From above it is clear given complex number lies on \(\mathrm{Y}\)-axis.
