NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
117698
If \(n\) is a positive integer greater than unity and \(Z\) is a complex number satisfying the equation \(Z^{\mathbf{n}}=(\mathbf{1}+\mathbf{Z})^{\mathbf{n}}\) then
1 \(\operatorname{Re}(Z)>0\)
2 \(\operatorname{Re} Z=0\)
3 \(\operatorname{Re}(Z)=\operatorname{Im}(Z)\)
4 \(\operatorname{Re}(Z)\lt 0\)
Explanation:
Exp: (D) Given, \(z\) is a complex number \& satisfying the equation - \(\mathrm{z}^{\mathrm{n}}=(1+\mathrm{z})^{\mathrm{n}}\) Then, \(\left(\frac{\mathrm{z}}{1+\mathrm{z}}\right)^{\mathrm{n}}=1\) \(\frac{\mathrm{z}}{1+\mathrm{z}}=1^{1 / \mathrm{n}}\) \(\because \quad \frac{\mathrm{z}}{1+\mathrm{z}}\) is a \(\mathrm{n}^{\text {th }}\) root of unity \(\therefore \quad\left|\frac{\mathrm{z}}{\mathrm{z}+1}\right|=1\) \(|z|=|z+1|\) \(\sqrt{x^2+y^2}=\sqrt{(x+1)^2+y^2} \quad\{\text { Let } z=x+i y\}\) squaring both side \(x^2+y^2=(x+1)^2+y^2\) \(2 x+1=0\) \(x=\frac{-1}{2}\) So, from above value \(\operatorname{Re}(\mathrm{z})\lt 0\)
SRM JEEE-2009
Complex Numbers and Quadratic Equation
117699
If \(z(\neq-1)\) is a complex number such that \(\frac{z-1}{z+1}\) is purely imaginary, then \(|z|\) is
1 1
2 2
3 3
4 5
Explanation:
A Let, \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) As, \(\frac{z-1}{z+1}\) is purely imaginary \(\operatorname{Re}\left(\frac{z-1}{z+1}\right)=0\) \(\operatorname{Re}\left(\frac{x+i y-1}{x+i y+1}\right)=0\) \(\operatorname{Re}\left[\frac{(x-1)+i y}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y}\right]=0\) \(\operatorname{Re}\left[\frac{(x-1)(x+1)-i(x-1) y+i y(x+1)+y^2}{(x+1)^2+y^2}\right]=0\) \(\operatorname{Re}\left[\frac{\left(x^2-1\right)+y^2+i\{y(x+1)-(x-1) y\}}{(x+1)^2+y^2}\right]=0\) \(\operatorname{Re}\left[\frac{x^2+y^2-1}{(x+1)^2+y^2}+i \frac{2 y}{(x+1)^2+y^2}\right]=0\) \(\frac{x^2+y^2-1}{(x+1)^2+y^2}=0\) \(x^2+y^2-1=0\) \(x^2+y^2=1\) \(|z|=1 \quad\left[z=x+i y \Rightarrow|z|=\sqrt{x^2+y^2}\right]\)
SRM JEEE-2011
Complex Numbers and Quadratic Equation
117700
If \(z=a+i b\) then \(i \log \left(\frac{\bar{z}}{z}\right)\) is equal to
117698
If \(n\) is a positive integer greater than unity and \(Z\) is a complex number satisfying the equation \(Z^{\mathbf{n}}=(\mathbf{1}+\mathbf{Z})^{\mathbf{n}}\) then
1 \(\operatorname{Re}(Z)>0\)
2 \(\operatorname{Re} Z=0\)
3 \(\operatorname{Re}(Z)=\operatorname{Im}(Z)\)
4 \(\operatorname{Re}(Z)\lt 0\)
Explanation:
Exp: (D) Given, \(z\) is a complex number \& satisfying the equation - \(\mathrm{z}^{\mathrm{n}}=(1+\mathrm{z})^{\mathrm{n}}\) Then, \(\left(\frac{\mathrm{z}}{1+\mathrm{z}}\right)^{\mathrm{n}}=1\) \(\frac{\mathrm{z}}{1+\mathrm{z}}=1^{1 / \mathrm{n}}\) \(\because \quad \frac{\mathrm{z}}{1+\mathrm{z}}\) is a \(\mathrm{n}^{\text {th }}\) root of unity \(\therefore \quad\left|\frac{\mathrm{z}}{\mathrm{z}+1}\right|=1\) \(|z|=|z+1|\) \(\sqrt{x^2+y^2}=\sqrt{(x+1)^2+y^2} \quad\{\text { Let } z=x+i y\}\) squaring both side \(x^2+y^2=(x+1)^2+y^2\) \(2 x+1=0\) \(x=\frac{-1}{2}\) So, from above value \(\operatorname{Re}(\mathrm{z})\lt 0\)
SRM JEEE-2009
Complex Numbers and Quadratic Equation
117699
If \(z(\neq-1)\) is a complex number such that \(\frac{z-1}{z+1}\) is purely imaginary, then \(|z|\) is
1 1
2 2
3 3
4 5
Explanation:
A Let, \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) As, \(\frac{z-1}{z+1}\) is purely imaginary \(\operatorname{Re}\left(\frac{z-1}{z+1}\right)=0\) \(\operatorname{Re}\left(\frac{x+i y-1}{x+i y+1}\right)=0\) \(\operatorname{Re}\left[\frac{(x-1)+i y}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y}\right]=0\) \(\operatorname{Re}\left[\frac{(x-1)(x+1)-i(x-1) y+i y(x+1)+y^2}{(x+1)^2+y^2}\right]=0\) \(\operatorname{Re}\left[\frac{\left(x^2-1\right)+y^2+i\{y(x+1)-(x-1) y\}}{(x+1)^2+y^2}\right]=0\) \(\operatorname{Re}\left[\frac{x^2+y^2-1}{(x+1)^2+y^2}+i \frac{2 y}{(x+1)^2+y^2}\right]=0\) \(\frac{x^2+y^2-1}{(x+1)^2+y^2}=0\) \(x^2+y^2-1=0\) \(x^2+y^2=1\) \(|z|=1 \quad\left[z=x+i y \Rightarrow|z|=\sqrt{x^2+y^2}\right]\)
SRM JEEE-2011
Complex Numbers and Quadratic Equation
117700
If \(z=a+i b\) then \(i \log \left(\frac{\bar{z}}{z}\right)\) is equal to
117698
If \(n\) is a positive integer greater than unity and \(Z\) is a complex number satisfying the equation \(Z^{\mathbf{n}}=(\mathbf{1}+\mathbf{Z})^{\mathbf{n}}\) then
1 \(\operatorname{Re}(Z)>0\)
2 \(\operatorname{Re} Z=0\)
3 \(\operatorname{Re}(Z)=\operatorname{Im}(Z)\)
4 \(\operatorname{Re}(Z)\lt 0\)
Explanation:
Exp: (D) Given, \(z\) is a complex number \& satisfying the equation - \(\mathrm{z}^{\mathrm{n}}=(1+\mathrm{z})^{\mathrm{n}}\) Then, \(\left(\frac{\mathrm{z}}{1+\mathrm{z}}\right)^{\mathrm{n}}=1\) \(\frac{\mathrm{z}}{1+\mathrm{z}}=1^{1 / \mathrm{n}}\) \(\because \quad \frac{\mathrm{z}}{1+\mathrm{z}}\) is a \(\mathrm{n}^{\text {th }}\) root of unity \(\therefore \quad\left|\frac{\mathrm{z}}{\mathrm{z}+1}\right|=1\) \(|z|=|z+1|\) \(\sqrt{x^2+y^2}=\sqrt{(x+1)^2+y^2} \quad\{\text { Let } z=x+i y\}\) squaring both side \(x^2+y^2=(x+1)^2+y^2\) \(2 x+1=0\) \(x=\frac{-1}{2}\) So, from above value \(\operatorname{Re}(\mathrm{z})\lt 0\)
SRM JEEE-2009
Complex Numbers and Quadratic Equation
117699
If \(z(\neq-1)\) is a complex number such that \(\frac{z-1}{z+1}\) is purely imaginary, then \(|z|\) is
1 1
2 2
3 3
4 5
Explanation:
A Let, \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) As, \(\frac{z-1}{z+1}\) is purely imaginary \(\operatorname{Re}\left(\frac{z-1}{z+1}\right)=0\) \(\operatorname{Re}\left(\frac{x+i y-1}{x+i y+1}\right)=0\) \(\operatorname{Re}\left[\frac{(x-1)+i y}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y}\right]=0\) \(\operatorname{Re}\left[\frac{(x-1)(x+1)-i(x-1) y+i y(x+1)+y^2}{(x+1)^2+y^2}\right]=0\) \(\operatorname{Re}\left[\frac{\left(x^2-1\right)+y^2+i\{y(x+1)-(x-1) y\}}{(x+1)^2+y^2}\right]=0\) \(\operatorname{Re}\left[\frac{x^2+y^2-1}{(x+1)^2+y^2}+i \frac{2 y}{(x+1)^2+y^2}\right]=0\) \(\frac{x^2+y^2-1}{(x+1)^2+y^2}=0\) \(x^2+y^2-1=0\) \(x^2+y^2=1\) \(|z|=1 \quad\left[z=x+i y \Rightarrow|z|=\sqrt{x^2+y^2}\right]\)
SRM JEEE-2011
Complex Numbers and Quadratic Equation
117700
If \(z=a+i b\) then \(i \log \left(\frac{\bar{z}}{z}\right)\) is equal to
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Complex Numbers and Quadratic Equation
117698
If \(n\) is a positive integer greater than unity and \(Z\) is a complex number satisfying the equation \(Z^{\mathbf{n}}=(\mathbf{1}+\mathbf{Z})^{\mathbf{n}}\) then
1 \(\operatorname{Re}(Z)>0\)
2 \(\operatorname{Re} Z=0\)
3 \(\operatorname{Re}(Z)=\operatorname{Im}(Z)\)
4 \(\operatorname{Re}(Z)\lt 0\)
Explanation:
Exp: (D) Given, \(z\) is a complex number \& satisfying the equation - \(\mathrm{z}^{\mathrm{n}}=(1+\mathrm{z})^{\mathrm{n}}\) Then, \(\left(\frac{\mathrm{z}}{1+\mathrm{z}}\right)^{\mathrm{n}}=1\) \(\frac{\mathrm{z}}{1+\mathrm{z}}=1^{1 / \mathrm{n}}\) \(\because \quad \frac{\mathrm{z}}{1+\mathrm{z}}\) is a \(\mathrm{n}^{\text {th }}\) root of unity \(\therefore \quad\left|\frac{\mathrm{z}}{\mathrm{z}+1}\right|=1\) \(|z|=|z+1|\) \(\sqrt{x^2+y^2}=\sqrt{(x+1)^2+y^2} \quad\{\text { Let } z=x+i y\}\) squaring both side \(x^2+y^2=(x+1)^2+y^2\) \(2 x+1=0\) \(x=\frac{-1}{2}\) So, from above value \(\operatorname{Re}(\mathrm{z})\lt 0\)
SRM JEEE-2009
Complex Numbers and Quadratic Equation
117699
If \(z(\neq-1)\) is a complex number such that \(\frac{z-1}{z+1}\) is purely imaginary, then \(|z|\) is
1 1
2 2
3 3
4 5
Explanation:
A Let, \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) As, \(\frac{z-1}{z+1}\) is purely imaginary \(\operatorname{Re}\left(\frac{z-1}{z+1}\right)=0\) \(\operatorname{Re}\left(\frac{x+i y-1}{x+i y+1}\right)=0\) \(\operatorname{Re}\left[\frac{(x-1)+i y}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y}\right]=0\) \(\operatorname{Re}\left[\frac{(x-1)(x+1)-i(x-1) y+i y(x+1)+y^2}{(x+1)^2+y^2}\right]=0\) \(\operatorname{Re}\left[\frac{\left(x^2-1\right)+y^2+i\{y(x+1)-(x-1) y\}}{(x+1)^2+y^2}\right]=0\) \(\operatorname{Re}\left[\frac{x^2+y^2-1}{(x+1)^2+y^2}+i \frac{2 y}{(x+1)^2+y^2}\right]=0\) \(\frac{x^2+y^2-1}{(x+1)^2+y^2}=0\) \(x^2+y^2-1=0\) \(x^2+y^2=1\) \(|z|=1 \quad\left[z=x+i y \Rightarrow|z|=\sqrt{x^2+y^2}\right]\)
SRM JEEE-2011
Complex Numbers and Quadratic Equation
117700
If \(z=a+i b\) then \(i \log \left(\frac{\bar{z}}{z}\right)\) is equal to