117694
If \(\left(\frac{1+i}{1-i}\right)^x=1\) then
1 \(x=4 n+1 ; n \in N\)
2 \(x=2 n+1 ; n \in N\)
3 \(x=2 n ; n \in N\)
4 \(x=4 n ; n \in N\)
Explanation:
D Given, \(\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{x}}=1\) Now, Rationalizing the denominator \(\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^x=1\) \({\left[\frac{(1+i)^2}{1-i^2}\right]^x=1 \quad\left\{\because i^2=-1\right\}}\) \({\left[\frac{1+i^2+2 i}{1+1}\right]^x=1}\) \({\left[\frac{2 i}{2}\right]^x=1}\) \(i^x=1\) Thus, from above equation \(\mathrm{x}\) will be multiple of 4 \(\therefore \mathrm{x}=4 \mathrm{n}\), where \(\mathrm{n} \in \mathrm{N}\).
Karnataka CET-2021
Complex Numbers and Quadratic Equation
117695
If \((x+i y)^{1 / 3}=a+i b\), then \(\frac{x}{a}+\frac{y}{b}=\)
1 \(\mathrm{ab}\)
2 \(4 \mathrm{ab}\)
3 \(4\left(a^2-b^2\right)\)
4 \(4\left(a^2+b^2\right)\)
Explanation:
C Given, \((x+i y)^{1 / 3}=a+i b\) taking cube on both side of above equation \((x+i y)=(a+i b)^3\) \((x+i y)=a^3+i^3 b^3+3 a^2 i b+3 a i^2 b^2\) \((x+i y)=a^3-i b^3+3 a^2 i b-3 a b^2\) \((x+i y)=a^3-3 a b^2+i\left(3 a^2 b-b^3\right)\) After comparing real \& imaginary parts \(x=a\left(a^2-3 b^2\right) \& y=b\left(3 a^2-b^2\right)\) Now, \(\frac{x}{a}+\frac{y}{b}=a^2-3 b^2+3 a^2-b^2\) \(\frac{x}{a}+\frac{y}{b}=4 a^2-4 b^2\) \(\frac{x}{a}+\frac{y}{b}=4\left(a^2-b^2\right) .\)
Karnataka CET-2000
Complex Numbers and Quadratic Equation
117696
Let \(z_1\) and \(z_2\) be nth roots of unity which subtend a right angle at the origin. Then \(n\) must be of the form :
1 \(4 \mathrm{k}+3\)
2 \(4 \mathrm{k}\)
3 \(4 \mathrm{k}+3\)
4 \(4 \mathrm{k}+2\)
Explanation:
Exp: (B): \(\text { Given } \mathrm{z}_1 \text { and } \mathrm{z}_2 \text { be } \mathrm{n}^{\mathrm{th}} \text { roots of unity. }\) \(\therefore \mathrm{z}_1^{\mathrm{n}}=\mathrm{z}_2^{\mathrm{n}}=1\) \(\text { and } \mathrm{z}_1, \mathrm{z}_2 \text { subtend a right angle }\) \(\therefore \arg \left(\frac{\mathrm{z}_1}{\mathrm{z}_2}\right)=\frac{\pi}{2}\) \(\frac{\mathrm{z}_1}{\mathrm{z}_2}=\cos \frac{\pi}{2}+\mathrm{i} \sin \frac{\pi}{2}\) \(\frac{\mathrm{z}_1}{\mathrm{z}_2}=0+\mathrm{i}\) \(\text { Now, }\left(\frac{\mathrm{z}_1}{\mathrm{z}_2}\right)^{\mathrm{n}}=(\mathrm{i})^{\mathrm{n}}\) \(\mathrm{z}_1^{\mathrm{n}}=\mathrm{z}_2^{\mathrm{n}}=1\) \(\text { Thus, } \mathrm{i}^{\mathrm{n}}=1 \text { is true only when } \mathrm{n} \text { is the form of } 4 \mathrm{k} .\) Now, \(\left(\frac{z_1}{z_2}\right)^{\mathrm{n}}=(\mathrm{i})^{\mathrm{n}}\) Thus, \(\mathrm{i}^{\mathrm{n}}=1\) is true only when \(\mathrm{n}\) is the form of \(4 \mathrm{k}\).
Karnataka CET-2002
Complex Numbers and Quadratic Equation
117697
If the imaginary part of \(\frac{2+i}{a i-1}\) is zero, where a is a real number, then the value of a is equal to
1 \(\frac{1}{2}\)
2 2
3 \(-\frac{1}{2}\)
4 -2
Explanation:
C Given, \(\operatorname{Im}\left(\frac{2+i}{a i-1}\right)=0\) let \(\mathrm{z}=\frac{2+\mathrm{i}}{\mathrm{ai}-1}\) Now, Rationalizing the denominator \(z=\frac{2+i}{a i-1}=\frac{2+i}{-1+a i} \times\left(\frac{-1-a i}{-1-a i}\right)\) \(=\frac{-2-2 a i-i-a^2}{(-1)^2-(a i)^2}\) \(=\frac{(-2+a)-i(2 a+1)}{1+a^2}\) \(z=\frac{(-2+a)}{1+a^2}-\frac{i(2 a+1)}{1+a^2}\) \(\because \quad \operatorname{Im}(\mathrm{z})=0\) \(\therefore \quad \frac{-(2 \mathrm{a}+1)}{1+\mathrm{a}^2}=0\) \(2 \mathrm{a}+1=0\) \(\mathrm{a}=-1 / 2\)
117694
If \(\left(\frac{1+i}{1-i}\right)^x=1\) then
1 \(x=4 n+1 ; n \in N\)
2 \(x=2 n+1 ; n \in N\)
3 \(x=2 n ; n \in N\)
4 \(x=4 n ; n \in N\)
Explanation:
D Given, \(\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{x}}=1\) Now, Rationalizing the denominator \(\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^x=1\) \({\left[\frac{(1+i)^2}{1-i^2}\right]^x=1 \quad\left\{\because i^2=-1\right\}}\) \({\left[\frac{1+i^2+2 i}{1+1}\right]^x=1}\) \({\left[\frac{2 i}{2}\right]^x=1}\) \(i^x=1\) Thus, from above equation \(\mathrm{x}\) will be multiple of 4 \(\therefore \mathrm{x}=4 \mathrm{n}\), where \(\mathrm{n} \in \mathrm{N}\).
Karnataka CET-2021
Complex Numbers and Quadratic Equation
117695
If \((x+i y)^{1 / 3}=a+i b\), then \(\frac{x}{a}+\frac{y}{b}=\)
1 \(\mathrm{ab}\)
2 \(4 \mathrm{ab}\)
3 \(4\left(a^2-b^2\right)\)
4 \(4\left(a^2+b^2\right)\)
Explanation:
C Given, \((x+i y)^{1 / 3}=a+i b\) taking cube on both side of above equation \((x+i y)=(a+i b)^3\) \((x+i y)=a^3+i^3 b^3+3 a^2 i b+3 a i^2 b^2\) \((x+i y)=a^3-i b^3+3 a^2 i b-3 a b^2\) \((x+i y)=a^3-3 a b^2+i\left(3 a^2 b-b^3\right)\) After comparing real \& imaginary parts \(x=a\left(a^2-3 b^2\right) \& y=b\left(3 a^2-b^2\right)\) Now, \(\frac{x}{a}+\frac{y}{b}=a^2-3 b^2+3 a^2-b^2\) \(\frac{x}{a}+\frac{y}{b}=4 a^2-4 b^2\) \(\frac{x}{a}+\frac{y}{b}=4\left(a^2-b^2\right) .\)
Karnataka CET-2000
Complex Numbers and Quadratic Equation
117696
Let \(z_1\) and \(z_2\) be nth roots of unity which subtend a right angle at the origin. Then \(n\) must be of the form :
1 \(4 \mathrm{k}+3\)
2 \(4 \mathrm{k}\)
3 \(4 \mathrm{k}+3\)
4 \(4 \mathrm{k}+2\)
Explanation:
Exp: (B): \(\text { Given } \mathrm{z}_1 \text { and } \mathrm{z}_2 \text { be } \mathrm{n}^{\mathrm{th}} \text { roots of unity. }\) \(\therefore \mathrm{z}_1^{\mathrm{n}}=\mathrm{z}_2^{\mathrm{n}}=1\) \(\text { and } \mathrm{z}_1, \mathrm{z}_2 \text { subtend a right angle }\) \(\therefore \arg \left(\frac{\mathrm{z}_1}{\mathrm{z}_2}\right)=\frac{\pi}{2}\) \(\frac{\mathrm{z}_1}{\mathrm{z}_2}=\cos \frac{\pi}{2}+\mathrm{i} \sin \frac{\pi}{2}\) \(\frac{\mathrm{z}_1}{\mathrm{z}_2}=0+\mathrm{i}\) \(\text { Now, }\left(\frac{\mathrm{z}_1}{\mathrm{z}_2}\right)^{\mathrm{n}}=(\mathrm{i})^{\mathrm{n}}\) \(\mathrm{z}_1^{\mathrm{n}}=\mathrm{z}_2^{\mathrm{n}}=1\) \(\text { Thus, } \mathrm{i}^{\mathrm{n}}=1 \text { is true only when } \mathrm{n} \text { is the form of } 4 \mathrm{k} .\) Now, \(\left(\frac{z_1}{z_2}\right)^{\mathrm{n}}=(\mathrm{i})^{\mathrm{n}}\) Thus, \(\mathrm{i}^{\mathrm{n}}=1\) is true only when \(\mathrm{n}\) is the form of \(4 \mathrm{k}\).
Karnataka CET-2002
Complex Numbers and Quadratic Equation
117697
If the imaginary part of \(\frac{2+i}{a i-1}\) is zero, where a is a real number, then the value of a is equal to
1 \(\frac{1}{2}\)
2 2
3 \(-\frac{1}{2}\)
4 -2
Explanation:
C Given, \(\operatorname{Im}\left(\frac{2+i}{a i-1}\right)=0\) let \(\mathrm{z}=\frac{2+\mathrm{i}}{\mathrm{ai}-1}\) Now, Rationalizing the denominator \(z=\frac{2+i}{a i-1}=\frac{2+i}{-1+a i} \times\left(\frac{-1-a i}{-1-a i}\right)\) \(=\frac{-2-2 a i-i-a^2}{(-1)^2-(a i)^2}\) \(=\frac{(-2+a)-i(2 a+1)}{1+a^2}\) \(z=\frac{(-2+a)}{1+a^2}-\frac{i(2 a+1)}{1+a^2}\) \(\because \quad \operatorname{Im}(\mathrm{z})=0\) \(\therefore \quad \frac{-(2 \mathrm{a}+1)}{1+\mathrm{a}^2}=0\) \(2 \mathrm{a}+1=0\) \(\mathrm{a}=-1 / 2\)
117694
If \(\left(\frac{1+i}{1-i}\right)^x=1\) then
1 \(x=4 n+1 ; n \in N\)
2 \(x=2 n+1 ; n \in N\)
3 \(x=2 n ; n \in N\)
4 \(x=4 n ; n \in N\)
Explanation:
D Given, \(\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{x}}=1\) Now, Rationalizing the denominator \(\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^x=1\) \({\left[\frac{(1+i)^2}{1-i^2}\right]^x=1 \quad\left\{\because i^2=-1\right\}}\) \({\left[\frac{1+i^2+2 i}{1+1}\right]^x=1}\) \({\left[\frac{2 i}{2}\right]^x=1}\) \(i^x=1\) Thus, from above equation \(\mathrm{x}\) will be multiple of 4 \(\therefore \mathrm{x}=4 \mathrm{n}\), where \(\mathrm{n} \in \mathrm{N}\).
Karnataka CET-2021
Complex Numbers and Quadratic Equation
117695
If \((x+i y)^{1 / 3}=a+i b\), then \(\frac{x}{a}+\frac{y}{b}=\)
1 \(\mathrm{ab}\)
2 \(4 \mathrm{ab}\)
3 \(4\left(a^2-b^2\right)\)
4 \(4\left(a^2+b^2\right)\)
Explanation:
C Given, \((x+i y)^{1 / 3}=a+i b\) taking cube on both side of above equation \((x+i y)=(a+i b)^3\) \((x+i y)=a^3+i^3 b^3+3 a^2 i b+3 a i^2 b^2\) \((x+i y)=a^3-i b^3+3 a^2 i b-3 a b^2\) \((x+i y)=a^3-3 a b^2+i\left(3 a^2 b-b^3\right)\) After comparing real \& imaginary parts \(x=a\left(a^2-3 b^2\right) \& y=b\left(3 a^2-b^2\right)\) Now, \(\frac{x}{a}+\frac{y}{b}=a^2-3 b^2+3 a^2-b^2\) \(\frac{x}{a}+\frac{y}{b}=4 a^2-4 b^2\) \(\frac{x}{a}+\frac{y}{b}=4\left(a^2-b^2\right) .\)
Karnataka CET-2000
Complex Numbers and Quadratic Equation
117696
Let \(z_1\) and \(z_2\) be nth roots of unity which subtend a right angle at the origin. Then \(n\) must be of the form :
1 \(4 \mathrm{k}+3\)
2 \(4 \mathrm{k}\)
3 \(4 \mathrm{k}+3\)
4 \(4 \mathrm{k}+2\)
Explanation:
Exp: (B): \(\text { Given } \mathrm{z}_1 \text { and } \mathrm{z}_2 \text { be } \mathrm{n}^{\mathrm{th}} \text { roots of unity. }\) \(\therefore \mathrm{z}_1^{\mathrm{n}}=\mathrm{z}_2^{\mathrm{n}}=1\) \(\text { and } \mathrm{z}_1, \mathrm{z}_2 \text { subtend a right angle }\) \(\therefore \arg \left(\frac{\mathrm{z}_1}{\mathrm{z}_2}\right)=\frac{\pi}{2}\) \(\frac{\mathrm{z}_1}{\mathrm{z}_2}=\cos \frac{\pi}{2}+\mathrm{i} \sin \frac{\pi}{2}\) \(\frac{\mathrm{z}_1}{\mathrm{z}_2}=0+\mathrm{i}\) \(\text { Now, }\left(\frac{\mathrm{z}_1}{\mathrm{z}_2}\right)^{\mathrm{n}}=(\mathrm{i})^{\mathrm{n}}\) \(\mathrm{z}_1^{\mathrm{n}}=\mathrm{z}_2^{\mathrm{n}}=1\) \(\text { Thus, } \mathrm{i}^{\mathrm{n}}=1 \text { is true only when } \mathrm{n} \text { is the form of } 4 \mathrm{k} .\) Now, \(\left(\frac{z_1}{z_2}\right)^{\mathrm{n}}=(\mathrm{i})^{\mathrm{n}}\) Thus, \(\mathrm{i}^{\mathrm{n}}=1\) is true only when \(\mathrm{n}\) is the form of \(4 \mathrm{k}\).
Karnataka CET-2002
Complex Numbers and Quadratic Equation
117697
If the imaginary part of \(\frac{2+i}{a i-1}\) is zero, where a is a real number, then the value of a is equal to
1 \(\frac{1}{2}\)
2 2
3 \(-\frac{1}{2}\)
4 -2
Explanation:
C Given, \(\operatorname{Im}\left(\frac{2+i}{a i-1}\right)=0\) let \(\mathrm{z}=\frac{2+\mathrm{i}}{\mathrm{ai}-1}\) Now, Rationalizing the denominator \(z=\frac{2+i}{a i-1}=\frac{2+i}{-1+a i} \times\left(\frac{-1-a i}{-1-a i}\right)\) \(=\frac{-2-2 a i-i-a^2}{(-1)^2-(a i)^2}\) \(=\frac{(-2+a)-i(2 a+1)}{1+a^2}\) \(z=\frac{(-2+a)}{1+a^2}-\frac{i(2 a+1)}{1+a^2}\) \(\because \quad \operatorname{Im}(\mathrm{z})=0\) \(\therefore \quad \frac{-(2 \mathrm{a}+1)}{1+\mathrm{a}^2}=0\) \(2 \mathrm{a}+1=0\) \(\mathrm{a}=-1 / 2\)
117694
If \(\left(\frac{1+i}{1-i}\right)^x=1\) then
1 \(x=4 n+1 ; n \in N\)
2 \(x=2 n+1 ; n \in N\)
3 \(x=2 n ; n \in N\)
4 \(x=4 n ; n \in N\)
Explanation:
D Given, \(\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{x}}=1\) Now, Rationalizing the denominator \(\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^x=1\) \({\left[\frac{(1+i)^2}{1-i^2}\right]^x=1 \quad\left\{\because i^2=-1\right\}}\) \({\left[\frac{1+i^2+2 i}{1+1}\right]^x=1}\) \({\left[\frac{2 i}{2}\right]^x=1}\) \(i^x=1\) Thus, from above equation \(\mathrm{x}\) will be multiple of 4 \(\therefore \mathrm{x}=4 \mathrm{n}\), where \(\mathrm{n} \in \mathrm{N}\).
Karnataka CET-2021
Complex Numbers and Quadratic Equation
117695
If \((x+i y)^{1 / 3}=a+i b\), then \(\frac{x}{a}+\frac{y}{b}=\)
1 \(\mathrm{ab}\)
2 \(4 \mathrm{ab}\)
3 \(4\left(a^2-b^2\right)\)
4 \(4\left(a^2+b^2\right)\)
Explanation:
C Given, \((x+i y)^{1 / 3}=a+i b\) taking cube on both side of above equation \((x+i y)=(a+i b)^3\) \((x+i y)=a^3+i^3 b^3+3 a^2 i b+3 a i^2 b^2\) \((x+i y)=a^3-i b^3+3 a^2 i b-3 a b^2\) \((x+i y)=a^3-3 a b^2+i\left(3 a^2 b-b^3\right)\) After comparing real \& imaginary parts \(x=a\left(a^2-3 b^2\right) \& y=b\left(3 a^2-b^2\right)\) Now, \(\frac{x}{a}+\frac{y}{b}=a^2-3 b^2+3 a^2-b^2\) \(\frac{x}{a}+\frac{y}{b}=4 a^2-4 b^2\) \(\frac{x}{a}+\frac{y}{b}=4\left(a^2-b^2\right) .\)
Karnataka CET-2000
Complex Numbers and Quadratic Equation
117696
Let \(z_1\) and \(z_2\) be nth roots of unity which subtend a right angle at the origin. Then \(n\) must be of the form :
1 \(4 \mathrm{k}+3\)
2 \(4 \mathrm{k}\)
3 \(4 \mathrm{k}+3\)
4 \(4 \mathrm{k}+2\)
Explanation:
Exp: (B): \(\text { Given } \mathrm{z}_1 \text { and } \mathrm{z}_2 \text { be } \mathrm{n}^{\mathrm{th}} \text { roots of unity. }\) \(\therefore \mathrm{z}_1^{\mathrm{n}}=\mathrm{z}_2^{\mathrm{n}}=1\) \(\text { and } \mathrm{z}_1, \mathrm{z}_2 \text { subtend a right angle }\) \(\therefore \arg \left(\frac{\mathrm{z}_1}{\mathrm{z}_2}\right)=\frac{\pi}{2}\) \(\frac{\mathrm{z}_1}{\mathrm{z}_2}=\cos \frac{\pi}{2}+\mathrm{i} \sin \frac{\pi}{2}\) \(\frac{\mathrm{z}_1}{\mathrm{z}_2}=0+\mathrm{i}\) \(\text { Now, }\left(\frac{\mathrm{z}_1}{\mathrm{z}_2}\right)^{\mathrm{n}}=(\mathrm{i})^{\mathrm{n}}\) \(\mathrm{z}_1^{\mathrm{n}}=\mathrm{z}_2^{\mathrm{n}}=1\) \(\text { Thus, } \mathrm{i}^{\mathrm{n}}=1 \text { is true only when } \mathrm{n} \text { is the form of } 4 \mathrm{k} .\) Now, \(\left(\frac{z_1}{z_2}\right)^{\mathrm{n}}=(\mathrm{i})^{\mathrm{n}}\) Thus, \(\mathrm{i}^{\mathrm{n}}=1\) is true only when \(\mathrm{n}\) is the form of \(4 \mathrm{k}\).
Karnataka CET-2002
Complex Numbers and Quadratic Equation
117697
If the imaginary part of \(\frac{2+i}{a i-1}\) is zero, where a is a real number, then the value of a is equal to
1 \(\frac{1}{2}\)
2 2
3 \(-\frac{1}{2}\)
4 -2
Explanation:
C Given, \(\operatorname{Im}\left(\frac{2+i}{a i-1}\right)=0\) let \(\mathrm{z}=\frac{2+\mathrm{i}}{\mathrm{ai}-1}\) Now, Rationalizing the denominator \(z=\frac{2+i}{a i-1}=\frac{2+i}{-1+a i} \times\left(\frac{-1-a i}{-1-a i}\right)\) \(=\frac{-2-2 a i-i-a^2}{(-1)^2-(a i)^2}\) \(=\frac{(-2+a)-i(2 a+1)}{1+a^2}\) \(z=\frac{(-2+a)}{1+a^2}-\frac{i(2 a+1)}{1+a^2}\) \(\because \quad \operatorname{Im}(\mathrm{z})=0\) \(\therefore \quad \frac{-(2 \mathrm{a}+1)}{1+\mathrm{a}^2}=0\) \(2 \mathrm{a}+1=0\) \(\mathrm{a}=-1 / 2\)
