NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
117817
If \(z=x+i y\), then the equation \(|z+1|=|z-1|\) represents
1 a parabola
2 \(x\)-axis
3 y-axis
4 a circle
Explanation:
C Given, \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(\text { then, } \quad|\mathrm{z}+1|=|\mathrm{z}-1|\) \(|\mathrm{x}+\mathrm{iy}+1|=|\mathrm{x}+\mathrm{iy}-1|\) \(|(\mathrm{x}+1)+\mathrm{iy}|=|(\mathrm{x}-1)+\mathrm{iy}|\) \(\sqrt{(\mathrm{x}+1)^2+\mathrm{y}^2}=\sqrt{(\mathrm{x}-1)^2+\mathrm{y}^2}\) \(\text { Squaring both side }\) Squaring both side \(x^2+1+2 x+y^2=x^2+1-2 x+y^2\) \(x^2+1+2 x+y^2-x^2-1+2 x-y^2\) \(4 x=0\) \(x=0 right.\)So, the locus of given equation is lies on \(y\) - axis
Karnataka CET-2020
Complex Numbers and Quadratic Equation
117818
If \(1, \omega, \omega^2\) are three cube roots of unity, then \(\left(1-\omega+\omega^2\right)\left(1+\omega-\omega^2\right)\) is
1 2
2 4
3 1
4 3
Explanation:
B If \(1, \omega, \omega^2\) are cube roots of unit, Then, \(1+\omega+\omega^2=0\) \(1+\omega^2=-\omega \& \omega^3=1\) Now, the value of \(\left(1-\omega+\omega^2\right)\left(1+\omega-\omega^2\right)\) \(=(-\omega-\omega)\left(-\omega^2-\omega^2\right)\left\{\because 1+\omega^2=-\omega \& 1+\omega=-\omega^2\right\}\) \(=-2 \omega\left(-2 \omega^2\right)\) \(=4 \omega^3\left\{\because \omega^3=1\right\}\) \(=4\)
Karnataka CET-2015
Complex Numbers and Quadratic Equation
117819
If \(z=\frac{(\sqrt{3}+i)^3(3 i+4)^2}{(8+6 i)^2}\), then \(|z|\) is equal to
117817
If \(z=x+i y\), then the equation \(|z+1|=|z-1|\) represents
1 a parabola
2 \(x\)-axis
3 y-axis
4 a circle
Explanation:
C Given, \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(\text { then, } \quad|\mathrm{z}+1|=|\mathrm{z}-1|\) \(|\mathrm{x}+\mathrm{iy}+1|=|\mathrm{x}+\mathrm{iy}-1|\) \(|(\mathrm{x}+1)+\mathrm{iy}|=|(\mathrm{x}-1)+\mathrm{iy}|\) \(\sqrt{(\mathrm{x}+1)^2+\mathrm{y}^2}=\sqrt{(\mathrm{x}-1)^2+\mathrm{y}^2}\) \(\text { Squaring both side }\) Squaring both side \(x^2+1+2 x+y^2=x^2+1-2 x+y^2\) \(x^2+1+2 x+y^2-x^2-1+2 x-y^2\) \(4 x=0\) \(x=0 right.\)So, the locus of given equation is lies on \(y\) - axis
Karnataka CET-2020
Complex Numbers and Quadratic Equation
117818
If \(1, \omega, \omega^2\) are three cube roots of unity, then \(\left(1-\omega+\omega^2\right)\left(1+\omega-\omega^2\right)\) is
1 2
2 4
3 1
4 3
Explanation:
B If \(1, \omega, \omega^2\) are cube roots of unit, Then, \(1+\omega+\omega^2=0\) \(1+\omega^2=-\omega \& \omega^3=1\) Now, the value of \(\left(1-\omega+\omega^2\right)\left(1+\omega-\omega^2\right)\) \(=(-\omega-\omega)\left(-\omega^2-\omega^2\right)\left\{\because 1+\omega^2=-\omega \& 1+\omega=-\omega^2\right\}\) \(=-2 \omega\left(-2 \omega^2\right)\) \(=4 \omega^3\left\{\because \omega^3=1\right\}\) \(=4\)
Karnataka CET-2015
Complex Numbers and Quadratic Equation
117819
If \(z=\frac{(\sqrt{3}+i)^3(3 i+4)^2}{(8+6 i)^2}\), then \(|z|\) is equal to
117817
If \(z=x+i y\), then the equation \(|z+1|=|z-1|\) represents
1 a parabola
2 \(x\)-axis
3 y-axis
4 a circle
Explanation:
C Given, \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(\text { then, } \quad|\mathrm{z}+1|=|\mathrm{z}-1|\) \(|\mathrm{x}+\mathrm{iy}+1|=|\mathrm{x}+\mathrm{iy}-1|\) \(|(\mathrm{x}+1)+\mathrm{iy}|=|(\mathrm{x}-1)+\mathrm{iy}|\) \(\sqrt{(\mathrm{x}+1)^2+\mathrm{y}^2}=\sqrt{(\mathrm{x}-1)^2+\mathrm{y}^2}\) \(\text { Squaring both side }\) Squaring both side \(x^2+1+2 x+y^2=x^2+1-2 x+y^2\) \(x^2+1+2 x+y^2-x^2-1+2 x-y^2\) \(4 x=0\) \(x=0 right.\)So, the locus of given equation is lies on \(y\) - axis
Karnataka CET-2020
Complex Numbers and Quadratic Equation
117818
If \(1, \omega, \omega^2\) are three cube roots of unity, then \(\left(1-\omega+\omega^2\right)\left(1+\omega-\omega^2\right)\) is
1 2
2 4
3 1
4 3
Explanation:
B If \(1, \omega, \omega^2\) are cube roots of unit, Then, \(1+\omega+\omega^2=0\) \(1+\omega^2=-\omega \& \omega^3=1\) Now, the value of \(\left(1-\omega+\omega^2\right)\left(1+\omega-\omega^2\right)\) \(=(-\omega-\omega)\left(-\omega^2-\omega^2\right)\left\{\because 1+\omega^2=-\omega \& 1+\omega=-\omega^2\right\}\) \(=-2 \omega\left(-2 \omega^2\right)\) \(=4 \omega^3\left\{\because \omega^3=1\right\}\) \(=4\)
Karnataka CET-2015
Complex Numbers and Quadratic Equation
117819
If \(z=\frac{(\sqrt{3}+i)^3(3 i+4)^2}{(8+6 i)^2}\), then \(|z|\) is equal to
117817
If \(z=x+i y\), then the equation \(|z+1|=|z-1|\) represents
1 a parabola
2 \(x\)-axis
3 y-axis
4 a circle
Explanation:
C Given, \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(\text { then, } \quad|\mathrm{z}+1|=|\mathrm{z}-1|\) \(|\mathrm{x}+\mathrm{iy}+1|=|\mathrm{x}+\mathrm{iy}-1|\) \(|(\mathrm{x}+1)+\mathrm{iy}|=|(\mathrm{x}-1)+\mathrm{iy}|\) \(\sqrt{(\mathrm{x}+1)^2+\mathrm{y}^2}=\sqrt{(\mathrm{x}-1)^2+\mathrm{y}^2}\) \(\text { Squaring both side }\) Squaring both side \(x^2+1+2 x+y^2=x^2+1-2 x+y^2\) \(x^2+1+2 x+y^2-x^2-1+2 x-y^2\) \(4 x=0\) \(x=0 right.\)So, the locus of given equation is lies on \(y\) - axis
Karnataka CET-2020
Complex Numbers and Quadratic Equation
117818
If \(1, \omega, \omega^2\) are three cube roots of unity, then \(\left(1-\omega+\omega^2\right)\left(1+\omega-\omega^2\right)\) is
1 2
2 4
3 1
4 3
Explanation:
B If \(1, \omega, \omega^2\) are cube roots of unit, Then, \(1+\omega+\omega^2=0\) \(1+\omega^2=-\omega \& \omega^3=1\) Now, the value of \(\left(1-\omega+\omega^2\right)\left(1+\omega-\omega^2\right)\) \(=(-\omega-\omega)\left(-\omega^2-\omega^2\right)\left\{\because 1+\omega^2=-\omega \& 1+\omega=-\omega^2\right\}\) \(=-2 \omega\left(-2 \omega^2\right)\) \(=4 \omega^3\left\{\because \omega^3=1\right\}\) \(=4\)
Karnataka CET-2015
Complex Numbers and Quadratic Equation
117819
If \(z=\frac{(\sqrt{3}+i)^3(3 i+4)^2}{(8+6 i)^2}\), then \(|z|\) is equal to
