117487
The simplified form of \(\mathbf{i}^{\mathrm{n}}+\mathrm{i}^{\mathrm{n}+1}+\mathrm{i}^{\mathrm{n}+2}+\mathrm{i}^{\mathrm{n}+3}\) is
1 0
2 1
3 -1
4 \(\mathrm{i}\)
Explanation:
A The simplified form of \(i^n+i^{n+1}+i^{n+2}+i^{n+3}\) is \(=\quad i^n+i^n \cdot i+i^n \cdot i^2+i^n \cdot i^3\) \(=\quad i^n\left(1+i+i^2+i^3\right)\) \(=\quad \mathrm{i}^{\mathrm{n}}(1+\mathrm{i}-1-\mathrm{i})\left\{\because \mathrm{i}^2=-1 \& \mathrm{i}^3=-\mathrm{i}\right\}\) \(=0\)
Karnataka CET-2016
Complex Numbers and Quadratic Equation
117488
If \(\left(\frac{1+i}{1+i}\right)^{\mathrm{m}}=1\), then the least positive integral value of \(\mathrm{m}\) is
1 4
2 1
3 2
4 3
Explanation:
A Given \(\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{m}}=1\) Now, Rationalizing the L.H.S of equation \(=\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{\mathrm{m}}=1\) \(=\left[\frac{(1+i)^2}{1^2-i^2}\right]^{\mathrm{m}}=1\) \(=\left[\frac{1+\mathrm{i}^2+2 \mathrm{i}}{1+1}\right]^{\mathrm{m}}=1\) \(=\left[\frac{2 i}{2}\right]^{\mathrm{m}}=1\) \(\mathrm{i}^{\mathrm{m}}=1 \ldots . . \text { (i) }\) we know that, \(\mathrm{i}^2=-1\) squaring both side \(\left(\mathrm{i}^2\right)^2=(-1)^2\) \(\mathrm{i}^4=1 \ldots . . \text { (ii) }\) comparing equation (i) \& (ii) \(\mathrm{m}=4\)
Karnataka CET-2017
Complex Numbers and Quadratic Equation
117489
The complex number \(\frac{1+2 i}{1-i}\) lies in
1 fourth quadrant
2 first quadrant
3 second quadrant
4 thrid quadrant
Explanation:
C Let, \(\mathrm{z}=\frac{1+2 \mathrm{i}}{1-\mathrm{i}}\) Multiplying denominator and numerator by \((1+i)\) \(=\frac{1+2 i}{1-i} \times \frac{(1+i)}{1+i}\) \(=\frac{1+i+2 i+2 i^2}{1-i^2}\) \(=\frac{1+3 i-2}{1+1}\) \(=\frac{-1+3 i}{2}=\frac{-1}{2}+i \frac{3}{2}\) \(\left\{\because \mathrm{i}^2=-1\right\}\) Comparing with \(\mathrm{x}+\) iy, we get real part is negative \& imaginary part is positive So, complex number \(\frac{1+2 \mathrm{i}}{1-\mathrm{i}}\) lies in second quadrant.
Karnataka CET-2009
Complex Numbers and Quadratic Equation
117490
The conjugate of the complex number \(\frac{(1+i)^2}{1-i}\) is
1 \(1-\mathrm{i}\)
2 \(1+\mathrm{i}\)
3 \(-1+i\)
4 \(-1-\mathrm{i}\)
Explanation:
D First of all simplify the given complex numbers \(z=\frac{(1+i)^2}{1-i}\) \(z=\frac{1+i^2+2 i}{1-i}\) \(z=\frac{2 i}{1-i}\) Now rationalizing the denominator \(=\frac{2 i}{1-i} \times \frac{1+i}{1+i}=\frac{2 i+2 i^2}{1-i^2}\) \(=\frac{2 i-2}{2}=i-1=-1+i\) So, conjugate of complex number is \(-1-i\)
117487
The simplified form of \(\mathbf{i}^{\mathrm{n}}+\mathrm{i}^{\mathrm{n}+1}+\mathrm{i}^{\mathrm{n}+2}+\mathrm{i}^{\mathrm{n}+3}\) is
1 0
2 1
3 -1
4 \(\mathrm{i}\)
Explanation:
A The simplified form of \(i^n+i^{n+1}+i^{n+2}+i^{n+3}\) is \(=\quad i^n+i^n \cdot i+i^n \cdot i^2+i^n \cdot i^3\) \(=\quad i^n\left(1+i+i^2+i^3\right)\) \(=\quad \mathrm{i}^{\mathrm{n}}(1+\mathrm{i}-1-\mathrm{i})\left\{\because \mathrm{i}^2=-1 \& \mathrm{i}^3=-\mathrm{i}\right\}\) \(=0\)
Karnataka CET-2016
Complex Numbers and Quadratic Equation
117488
If \(\left(\frac{1+i}{1+i}\right)^{\mathrm{m}}=1\), then the least positive integral value of \(\mathrm{m}\) is
1 4
2 1
3 2
4 3
Explanation:
A Given \(\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{m}}=1\) Now, Rationalizing the L.H.S of equation \(=\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{\mathrm{m}}=1\) \(=\left[\frac{(1+i)^2}{1^2-i^2}\right]^{\mathrm{m}}=1\) \(=\left[\frac{1+\mathrm{i}^2+2 \mathrm{i}}{1+1}\right]^{\mathrm{m}}=1\) \(=\left[\frac{2 i}{2}\right]^{\mathrm{m}}=1\) \(\mathrm{i}^{\mathrm{m}}=1 \ldots . . \text { (i) }\) we know that, \(\mathrm{i}^2=-1\) squaring both side \(\left(\mathrm{i}^2\right)^2=(-1)^2\) \(\mathrm{i}^4=1 \ldots . . \text { (ii) }\) comparing equation (i) \& (ii) \(\mathrm{m}=4\)
Karnataka CET-2017
Complex Numbers and Quadratic Equation
117489
The complex number \(\frac{1+2 i}{1-i}\) lies in
1 fourth quadrant
2 first quadrant
3 second quadrant
4 thrid quadrant
Explanation:
C Let, \(\mathrm{z}=\frac{1+2 \mathrm{i}}{1-\mathrm{i}}\) Multiplying denominator and numerator by \((1+i)\) \(=\frac{1+2 i}{1-i} \times \frac{(1+i)}{1+i}\) \(=\frac{1+i+2 i+2 i^2}{1-i^2}\) \(=\frac{1+3 i-2}{1+1}\) \(=\frac{-1+3 i}{2}=\frac{-1}{2}+i \frac{3}{2}\) \(\left\{\because \mathrm{i}^2=-1\right\}\) Comparing with \(\mathrm{x}+\) iy, we get real part is negative \& imaginary part is positive So, complex number \(\frac{1+2 \mathrm{i}}{1-\mathrm{i}}\) lies in second quadrant.
Karnataka CET-2009
Complex Numbers and Quadratic Equation
117490
The conjugate of the complex number \(\frac{(1+i)^2}{1-i}\) is
1 \(1-\mathrm{i}\)
2 \(1+\mathrm{i}\)
3 \(-1+i\)
4 \(-1-\mathrm{i}\)
Explanation:
D First of all simplify the given complex numbers \(z=\frac{(1+i)^2}{1-i}\) \(z=\frac{1+i^2+2 i}{1-i}\) \(z=\frac{2 i}{1-i}\) Now rationalizing the denominator \(=\frac{2 i}{1-i} \times \frac{1+i}{1+i}=\frac{2 i+2 i^2}{1-i^2}\) \(=\frac{2 i-2}{2}=i-1=-1+i\) So, conjugate of complex number is \(-1-i\)
117487
The simplified form of \(\mathbf{i}^{\mathrm{n}}+\mathrm{i}^{\mathrm{n}+1}+\mathrm{i}^{\mathrm{n}+2}+\mathrm{i}^{\mathrm{n}+3}\) is
1 0
2 1
3 -1
4 \(\mathrm{i}\)
Explanation:
A The simplified form of \(i^n+i^{n+1}+i^{n+2}+i^{n+3}\) is \(=\quad i^n+i^n \cdot i+i^n \cdot i^2+i^n \cdot i^3\) \(=\quad i^n\left(1+i+i^2+i^3\right)\) \(=\quad \mathrm{i}^{\mathrm{n}}(1+\mathrm{i}-1-\mathrm{i})\left\{\because \mathrm{i}^2=-1 \& \mathrm{i}^3=-\mathrm{i}\right\}\) \(=0\)
Karnataka CET-2016
Complex Numbers and Quadratic Equation
117488
If \(\left(\frac{1+i}{1+i}\right)^{\mathrm{m}}=1\), then the least positive integral value of \(\mathrm{m}\) is
1 4
2 1
3 2
4 3
Explanation:
A Given \(\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{m}}=1\) Now, Rationalizing the L.H.S of equation \(=\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{\mathrm{m}}=1\) \(=\left[\frac{(1+i)^2}{1^2-i^2}\right]^{\mathrm{m}}=1\) \(=\left[\frac{1+\mathrm{i}^2+2 \mathrm{i}}{1+1}\right]^{\mathrm{m}}=1\) \(=\left[\frac{2 i}{2}\right]^{\mathrm{m}}=1\) \(\mathrm{i}^{\mathrm{m}}=1 \ldots . . \text { (i) }\) we know that, \(\mathrm{i}^2=-1\) squaring both side \(\left(\mathrm{i}^2\right)^2=(-1)^2\) \(\mathrm{i}^4=1 \ldots . . \text { (ii) }\) comparing equation (i) \& (ii) \(\mathrm{m}=4\)
Karnataka CET-2017
Complex Numbers and Quadratic Equation
117489
The complex number \(\frac{1+2 i}{1-i}\) lies in
1 fourth quadrant
2 first quadrant
3 second quadrant
4 thrid quadrant
Explanation:
C Let, \(\mathrm{z}=\frac{1+2 \mathrm{i}}{1-\mathrm{i}}\) Multiplying denominator and numerator by \((1+i)\) \(=\frac{1+2 i}{1-i} \times \frac{(1+i)}{1+i}\) \(=\frac{1+i+2 i+2 i^2}{1-i^2}\) \(=\frac{1+3 i-2}{1+1}\) \(=\frac{-1+3 i}{2}=\frac{-1}{2}+i \frac{3}{2}\) \(\left\{\because \mathrm{i}^2=-1\right\}\) Comparing with \(\mathrm{x}+\) iy, we get real part is negative \& imaginary part is positive So, complex number \(\frac{1+2 \mathrm{i}}{1-\mathrm{i}}\) lies in second quadrant.
Karnataka CET-2009
Complex Numbers and Quadratic Equation
117490
The conjugate of the complex number \(\frac{(1+i)^2}{1-i}\) is
1 \(1-\mathrm{i}\)
2 \(1+\mathrm{i}\)
3 \(-1+i\)
4 \(-1-\mathrm{i}\)
Explanation:
D First of all simplify the given complex numbers \(z=\frac{(1+i)^2}{1-i}\) \(z=\frac{1+i^2+2 i}{1-i}\) \(z=\frac{2 i}{1-i}\) Now rationalizing the denominator \(=\frac{2 i}{1-i} \times \frac{1+i}{1+i}=\frac{2 i+2 i^2}{1-i^2}\) \(=\frac{2 i-2}{2}=i-1=-1+i\) So, conjugate of complex number is \(-1-i\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
117487
The simplified form of \(\mathbf{i}^{\mathrm{n}}+\mathrm{i}^{\mathrm{n}+1}+\mathrm{i}^{\mathrm{n}+2}+\mathrm{i}^{\mathrm{n}+3}\) is
1 0
2 1
3 -1
4 \(\mathrm{i}\)
Explanation:
A The simplified form of \(i^n+i^{n+1}+i^{n+2}+i^{n+3}\) is \(=\quad i^n+i^n \cdot i+i^n \cdot i^2+i^n \cdot i^3\) \(=\quad i^n\left(1+i+i^2+i^3\right)\) \(=\quad \mathrm{i}^{\mathrm{n}}(1+\mathrm{i}-1-\mathrm{i})\left\{\because \mathrm{i}^2=-1 \& \mathrm{i}^3=-\mathrm{i}\right\}\) \(=0\)
Karnataka CET-2016
Complex Numbers and Quadratic Equation
117488
If \(\left(\frac{1+i}{1+i}\right)^{\mathrm{m}}=1\), then the least positive integral value of \(\mathrm{m}\) is
1 4
2 1
3 2
4 3
Explanation:
A Given \(\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{m}}=1\) Now, Rationalizing the L.H.S of equation \(=\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{\mathrm{m}}=1\) \(=\left[\frac{(1+i)^2}{1^2-i^2}\right]^{\mathrm{m}}=1\) \(=\left[\frac{1+\mathrm{i}^2+2 \mathrm{i}}{1+1}\right]^{\mathrm{m}}=1\) \(=\left[\frac{2 i}{2}\right]^{\mathrm{m}}=1\) \(\mathrm{i}^{\mathrm{m}}=1 \ldots . . \text { (i) }\) we know that, \(\mathrm{i}^2=-1\) squaring both side \(\left(\mathrm{i}^2\right)^2=(-1)^2\) \(\mathrm{i}^4=1 \ldots . . \text { (ii) }\) comparing equation (i) \& (ii) \(\mathrm{m}=4\)
Karnataka CET-2017
Complex Numbers and Quadratic Equation
117489
The complex number \(\frac{1+2 i}{1-i}\) lies in
1 fourth quadrant
2 first quadrant
3 second quadrant
4 thrid quadrant
Explanation:
C Let, \(\mathrm{z}=\frac{1+2 \mathrm{i}}{1-\mathrm{i}}\) Multiplying denominator and numerator by \((1+i)\) \(=\frac{1+2 i}{1-i} \times \frac{(1+i)}{1+i}\) \(=\frac{1+i+2 i+2 i^2}{1-i^2}\) \(=\frac{1+3 i-2}{1+1}\) \(=\frac{-1+3 i}{2}=\frac{-1}{2}+i \frac{3}{2}\) \(\left\{\because \mathrm{i}^2=-1\right\}\) Comparing with \(\mathrm{x}+\) iy, we get real part is negative \& imaginary part is positive So, complex number \(\frac{1+2 \mathrm{i}}{1-\mathrm{i}}\) lies in second quadrant.
Karnataka CET-2009
Complex Numbers and Quadratic Equation
117490
The conjugate of the complex number \(\frac{(1+i)^2}{1-i}\) is
1 \(1-\mathrm{i}\)
2 \(1+\mathrm{i}\)
3 \(-1+i\)
4 \(-1-\mathrm{i}\)
Explanation:
D First of all simplify the given complex numbers \(z=\frac{(1+i)^2}{1-i}\) \(z=\frac{1+i^2+2 i}{1-i}\) \(z=\frac{2 i}{1-i}\) Now rationalizing the denominator \(=\frac{2 i}{1-i} \times \frac{1+i}{1+i}=\frac{2 i+2 i^2}{1-i^2}\) \(=\frac{2 i-2}{2}=i-1=-1+i\) So, conjugate of complex number is \(-1-i\)
