117828
If \(a>0, a \in R, z=a+2 i\) and \(z|z|-a z+1=0\) then
1 \(z\) is always a positive real number
2 \(z\) is always a negative real number
3 \(z\) is purely imaginary number
4 such a complex \(\mathrm{z}\) does not exist
Explanation:
D Given, \(z=a+2 i\) \(a>0 \& a \in R\) Then, \(\quad \mathrm{z}|\mathrm{z}|-\mathrm{az}+1=0\) \(a+2 i|a+2 i|-a(a+2 i)+1=0\) \((a+2 i)|a+2 i|-a(a+2 i)=-1\) \((a+2 i)[|a+2 i|-a]=-1\) \(\sqrt{\mathrm{a}^2+4}-\mathrm{a}=\frac{-1}{\mathrm{a}+2 \mathrm{i}}\) R.H.S has imaginary part \& real while L.H.S has real part only which is not possible. So, such a complex \(\mathrm{z}\) does not exist.
BITSAT-2011
Complex Numbers and Quadratic Equation
117829
If \(\frac{1-i \alpha}{1+i \alpha}=A+i B\), then \(A^2+B^2\) equals to
1 1
2 \(\alpha^2\)
3 -1
4 \(-\alpha^2\)
Explanation:
A Given, \(A+i B=\frac{1-i \alpha}{1+i \alpha}\) then conjugate of \(A+i B=A-i B\) \(A-i B=\frac{1+i \alpha}{1-i \alpha}\) \(\Rightarrow(A+i B)(A-i B)=\frac{(1-i \alpha)(1+i \alpha)}{(1+i \alpha)(1-i \alpha)}=1\) \(\Rightarrow A^2+B^2=1\)
BITSAT-2010
Complex Numbers and Quadratic Equation
117830
The complex number \(z=z+\) iy which satisfies the equation \(\left|\frac{z-3 i}{z+3 i}\right|=1\), lies on
1 the \(\mathrm{X}\)-axis
2 the straight line \(y=3\)
3 a circle passing through origin
4 None of the above
Explanation:
Exp: (A) Given, complex number \(z=x+\) iy then, equation - \(\left|\frac{z-3 i}{z+3 i}\right|=1\) \(\left|\frac{x+i y-3 i}{x+i y+3 i}\right|=1\) \(|x+i(y-3)|=|x+i(y+3)|\) \(\sqrt{x^2+(y-3)^2}=\sqrt{x^2+(y+3)^2}\) squaring both side \(x^2+(y-3)^2=x^2+(y+3)^2\) \(-6 y=6 y\) \(12 y=0\) \(y=0\)So, It's lies on X-axis.
117828
If \(a>0, a \in R, z=a+2 i\) and \(z|z|-a z+1=0\) then
1 \(z\) is always a positive real number
2 \(z\) is always a negative real number
3 \(z\) is purely imaginary number
4 such a complex \(\mathrm{z}\) does not exist
Explanation:
D Given, \(z=a+2 i\) \(a>0 \& a \in R\) Then, \(\quad \mathrm{z}|\mathrm{z}|-\mathrm{az}+1=0\) \(a+2 i|a+2 i|-a(a+2 i)+1=0\) \((a+2 i)|a+2 i|-a(a+2 i)=-1\) \((a+2 i)[|a+2 i|-a]=-1\) \(\sqrt{\mathrm{a}^2+4}-\mathrm{a}=\frac{-1}{\mathrm{a}+2 \mathrm{i}}\) R.H.S has imaginary part \& real while L.H.S has real part only which is not possible. So, such a complex \(\mathrm{z}\) does not exist.
BITSAT-2011
Complex Numbers and Quadratic Equation
117829
If \(\frac{1-i \alpha}{1+i \alpha}=A+i B\), then \(A^2+B^2\) equals to
1 1
2 \(\alpha^2\)
3 -1
4 \(-\alpha^2\)
Explanation:
A Given, \(A+i B=\frac{1-i \alpha}{1+i \alpha}\) then conjugate of \(A+i B=A-i B\) \(A-i B=\frac{1+i \alpha}{1-i \alpha}\) \(\Rightarrow(A+i B)(A-i B)=\frac{(1-i \alpha)(1+i \alpha)}{(1+i \alpha)(1-i \alpha)}=1\) \(\Rightarrow A^2+B^2=1\)
BITSAT-2010
Complex Numbers and Quadratic Equation
117830
The complex number \(z=z+\) iy which satisfies the equation \(\left|\frac{z-3 i}{z+3 i}\right|=1\), lies on
1 the \(\mathrm{X}\)-axis
2 the straight line \(y=3\)
3 a circle passing through origin
4 None of the above
Explanation:
Exp: (A) Given, complex number \(z=x+\) iy then, equation - \(\left|\frac{z-3 i}{z+3 i}\right|=1\) \(\left|\frac{x+i y-3 i}{x+i y+3 i}\right|=1\) \(|x+i(y-3)|=|x+i(y+3)|\) \(\sqrt{x^2+(y-3)^2}=\sqrt{x^2+(y+3)^2}\) squaring both side \(x^2+(y-3)^2=x^2+(y+3)^2\) \(-6 y=6 y\) \(12 y=0\) \(y=0\)So, It's lies on X-axis.
117828
If \(a>0, a \in R, z=a+2 i\) and \(z|z|-a z+1=0\) then
1 \(z\) is always a positive real number
2 \(z\) is always a negative real number
3 \(z\) is purely imaginary number
4 such a complex \(\mathrm{z}\) does not exist
Explanation:
D Given, \(z=a+2 i\) \(a>0 \& a \in R\) Then, \(\quad \mathrm{z}|\mathrm{z}|-\mathrm{az}+1=0\) \(a+2 i|a+2 i|-a(a+2 i)+1=0\) \((a+2 i)|a+2 i|-a(a+2 i)=-1\) \((a+2 i)[|a+2 i|-a]=-1\) \(\sqrt{\mathrm{a}^2+4}-\mathrm{a}=\frac{-1}{\mathrm{a}+2 \mathrm{i}}\) R.H.S has imaginary part \& real while L.H.S has real part only which is not possible. So, such a complex \(\mathrm{z}\) does not exist.
BITSAT-2011
Complex Numbers and Quadratic Equation
117829
If \(\frac{1-i \alpha}{1+i \alpha}=A+i B\), then \(A^2+B^2\) equals to
1 1
2 \(\alpha^2\)
3 -1
4 \(-\alpha^2\)
Explanation:
A Given, \(A+i B=\frac{1-i \alpha}{1+i \alpha}\) then conjugate of \(A+i B=A-i B\) \(A-i B=\frac{1+i \alpha}{1-i \alpha}\) \(\Rightarrow(A+i B)(A-i B)=\frac{(1-i \alpha)(1+i \alpha)}{(1+i \alpha)(1-i \alpha)}=1\) \(\Rightarrow A^2+B^2=1\)
BITSAT-2010
Complex Numbers and Quadratic Equation
117830
The complex number \(z=z+\) iy which satisfies the equation \(\left|\frac{z-3 i}{z+3 i}\right|=1\), lies on
1 the \(\mathrm{X}\)-axis
2 the straight line \(y=3\)
3 a circle passing through origin
4 None of the above
Explanation:
Exp: (A) Given, complex number \(z=x+\) iy then, equation - \(\left|\frac{z-3 i}{z+3 i}\right|=1\) \(\left|\frac{x+i y-3 i}{x+i y+3 i}\right|=1\) \(|x+i(y-3)|=|x+i(y+3)|\) \(\sqrt{x^2+(y-3)^2}=\sqrt{x^2+(y+3)^2}\) squaring both side \(x^2+(y-3)^2=x^2+(y+3)^2\) \(-6 y=6 y\) \(12 y=0\) \(y=0\)So, It's lies on X-axis.
117828
If \(a>0, a \in R, z=a+2 i\) and \(z|z|-a z+1=0\) then
1 \(z\) is always a positive real number
2 \(z\) is always a negative real number
3 \(z\) is purely imaginary number
4 such a complex \(\mathrm{z}\) does not exist
Explanation:
D Given, \(z=a+2 i\) \(a>0 \& a \in R\) Then, \(\quad \mathrm{z}|\mathrm{z}|-\mathrm{az}+1=0\) \(a+2 i|a+2 i|-a(a+2 i)+1=0\) \((a+2 i)|a+2 i|-a(a+2 i)=-1\) \((a+2 i)[|a+2 i|-a]=-1\) \(\sqrt{\mathrm{a}^2+4}-\mathrm{a}=\frac{-1}{\mathrm{a}+2 \mathrm{i}}\) R.H.S has imaginary part \& real while L.H.S has real part only which is not possible. So, such a complex \(\mathrm{z}\) does not exist.
BITSAT-2011
Complex Numbers and Quadratic Equation
117829
If \(\frac{1-i \alpha}{1+i \alpha}=A+i B\), then \(A^2+B^2\) equals to
1 1
2 \(\alpha^2\)
3 -1
4 \(-\alpha^2\)
Explanation:
A Given, \(A+i B=\frac{1-i \alpha}{1+i \alpha}\) then conjugate of \(A+i B=A-i B\) \(A-i B=\frac{1+i \alpha}{1-i \alpha}\) \(\Rightarrow(A+i B)(A-i B)=\frac{(1-i \alpha)(1+i \alpha)}{(1+i \alpha)(1-i \alpha)}=1\) \(\Rightarrow A^2+B^2=1\)
BITSAT-2010
Complex Numbers and Quadratic Equation
117830
The complex number \(z=z+\) iy which satisfies the equation \(\left|\frac{z-3 i}{z+3 i}\right|=1\), lies on
1 the \(\mathrm{X}\)-axis
2 the straight line \(y=3\)
3 a circle passing through origin
4 None of the above
Explanation:
Exp: (A) Given, complex number \(z=x+\) iy then, equation - \(\left|\frac{z-3 i}{z+3 i}\right|=1\) \(\left|\frac{x+i y-3 i}{x+i y+3 i}\right|=1\) \(|x+i(y-3)|=|x+i(y+3)|\) \(\sqrt{x^2+(y-3)^2}=\sqrt{x^2+(y+3)^2}\) squaring both side \(x^2+(y-3)^2=x^2+(y+3)^2\) \(-6 y=6 y\) \(12 y=0\) \(y=0\)So, It's lies on X-axis.
