117702
The multiplicative inverse of \(\frac{3+4 i}{4-5 i}\) is
1 \(\left(\frac{-8}{25}, \frac{31}{25}\right)\)
2 \(\left(\frac{-8}{25}, \frac{-31}{25}\right)\)
3 \(\left(\frac{8}{25}, \frac{-31}{25}\right)\)
4 \(\left(\frac{8}{25}, \frac{31}{25}\right)\)
Explanation:
B Let \(z=\frac{3+4 i}{4-5 i}\) We have to calculate \(\mathrm{z}^{-1}\) i.e., \(\frac{1}{\mathrm{z}}\) \(\therefore z^{-1}=\frac{1}{z}=\frac{4-5 i}{3+4 i} \times \frac{3-4 i}{3-4 i}\) \(=\frac{12-15 i-16 i+20 i^2}{9-16 i^2}=\frac{12-31 i-20}{9-16(-1)}\) \(=\frac{-8-31 \mathrm{i}}{9+16}=\left(-\frac{8}{25}, \frac{-31}{25}\right)\)
COMEDK-2012
Complex Numbers and Quadratic Equation
117703
If a complex number lies in the III quadrant. Find the quadrant in which its conjugate lies.
1 I quadrant
2 II quadrant
3 III quadrant
4 IV quadrant
Explanation:
B Given, complex number lies in III quadrant. \(\therefore \quad \mathrm{z}=-\mathrm{a}-\mathrm{ib}\) \(\mathrm{a}\lt 0 \& \mathrm{~b}\lt 0\) So, it's conjugate is \(\bar{z}=-a+i b\) here \(\mathrm{a}\lt 0 \& \mathrm{~b}>0\) \(\therefore\) Conjugate lies in II quadrant.
COMEDK-2014
Complex Numbers and Quadratic Equation
117704
If \(z=r e^{i \theta}\), then \(\left|e^i\right|=\)
1 1
2 \(\mathrm{e}^{2 \mathrm{r} \sin \theta}\)
3 \(\mathrm{e}^{\mathrm{rsin} \theta}\)
4 \(\mathrm{e}^{-\mathrm{s} \sin \theta}\)
Explanation:
D Given, \(z=\mathrm{re}^{\mathrm{i} \theta}\) \(z=r(\cos \theta+i \sin \theta)\) \{From euller's formula \} Multiply by \(i\) to both side - \(\mathrm{iz}=\mathrm{ir}(\cos \theta+\mathrm{i} \sin \theta)\) \(\mathrm{iz}=\mathrm{ir} \cos \theta+\mathrm{i}^2 \mathrm{r} \sin \theta\) \(\mathrm{iz}=-\mathrm{r} \sin \theta+\mathrm{ir} \cos \theta\) \(\mathrm{e}^{\mathrm{iz}}=\mathrm{e}^{(-\mathrm{r} \sin \theta+\mathrm{ir} \cos \theta)}\) Taking modulus of both side \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right| .\left|\mathrm{e}^{\mathrm{ir} \cos \theta}\right|\) \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right| \times 1\left\{\because\left|\mathrm{e}^{\mathrm{i} \theta}\right|=1\right\}\) \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right|\)
COMEDK-2017
Complex Numbers and Quadratic Equation
117705
If the complex number \(z\) lies on a circle with centre at the origin and radius \(\frac{1}{4}\), then the complex number \(-1+8 \mathrm{z}\) lies on a circle with radius
1 4
2 1
3 3
4 2
Explanation:
D Given, \(\mathrm{z}\) is complex number lies on a circle with radius \(\frac{1}{4}\) we know that, \(|z|=r\) \(\therefore \quad|z| =\frac{1}{4}\) Let \(z^{\prime}=-1+8 z\) \(z=\frac{z^{\prime}+1}{8}\) Taking modules of both side \(|z|=\left|\frac{z^{\prime}+1}{8}\right|\) \(\frac{1}{4}=\left|\frac{z^{\prime}+1}{8}\right|\) \(\left|z^{\prime}+1\right|=2\) So, complex number \(z^{\prime}\) lies on circle with centre at \((-1\), \(0)\) and radius 2 .
VITEEE-2015
Complex Numbers and Quadratic Equation
117706
If the area of the triangle on the complex plane formed by the points \(z, z+i z\) and iz is 200 , then the value of \(3|z|\) must be equal to
1 20
2 40
3 60
4 80
Explanation:
C Let, \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(z+i z=x+i y+i(x+i y)\) \(=x-y+i x+i y \because i^2=-1\) \(=(x-y)+i(x+y)\) \(i z=i(x+i y) \because i^2=1\) \(=i x+i^2 y\) and \(\mathrm{iz} =\mathrm{i}(\mathrm{x}+\mathrm{iy})\) \(=\mathrm{ix}+\mathrm{i}^2 \mathrm{y} \because \mathrm{i}^2=1\) \(=-\mathrm{y}+\mathrm{ix}\) Then, the area of the triangle formed by these lines is \begin{align*} \begin{aligned} & \Delta=\frac{1}{2}\left\|\begin{array}{ccc} \mathrm{x} & \mathrm{y} & 1 \\ (\mathrm{x}-\mathrm{y}) & (\mathrm{x}+\mathrm{y}) & 1 \\ -\mathrm{y} & \mathrm{x} & 1 \end{array}\right\| \\ & \text { Applying } \mathrm{R}_2 \rightarrow \mathrm{R}_2-\left(\mathrm{R}_1+\mathrm{R}_3\right), \\ & \Delta=\frac{1}{2}\left\|\begin{array}{ccc} \mathrm{x} & \mathrm{y} & 1 \\ 0 & 0 & -1 \\ -\mathrm{y} & \mathrm{x} & 1 \end{array}\right\| \end{aligned} \end{align*} \(=\frac{1}{2}\left(\mathrm{x}^2+\mathrm{y}^2\right)\) \(\frac{1}{2}|z|^2=200\) \(|z|^2=400\) \(|z|=20\) \(\therefore 3|z|=3 \times 20\) \(3|z|=60\)
117702
The multiplicative inverse of \(\frac{3+4 i}{4-5 i}\) is
1 \(\left(\frac{-8}{25}, \frac{31}{25}\right)\)
2 \(\left(\frac{-8}{25}, \frac{-31}{25}\right)\)
3 \(\left(\frac{8}{25}, \frac{-31}{25}\right)\)
4 \(\left(\frac{8}{25}, \frac{31}{25}\right)\)
Explanation:
B Let \(z=\frac{3+4 i}{4-5 i}\) We have to calculate \(\mathrm{z}^{-1}\) i.e., \(\frac{1}{\mathrm{z}}\) \(\therefore z^{-1}=\frac{1}{z}=\frac{4-5 i}{3+4 i} \times \frac{3-4 i}{3-4 i}\) \(=\frac{12-15 i-16 i+20 i^2}{9-16 i^2}=\frac{12-31 i-20}{9-16(-1)}\) \(=\frac{-8-31 \mathrm{i}}{9+16}=\left(-\frac{8}{25}, \frac{-31}{25}\right)\)
COMEDK-2012
Complex Numbers and Quadratic Equation
117703
If a complex number lies in the III quadrant. Find the quadrant in which its conjugate lies.
1 I quadrant
2 II quadrant
3 III quadrant
4 IV quadrant
Explanation:
B Given, complex number lies in III quadrant. \(\therefore \quad \mathrm{z}=-\mathrm{a}-\mathrm{ib}\) \(\mathrm{a}\lt 0 \& \mathrm{~b}\lt 0\) So, it's conjugate is \(\bar{z}=-a+i b\) here \(\mathrm{a}\lt 0 \& \mathrm{~b}>0\) \(\therefore\) Conjugate lies in II quadrant.
COMEDK-2014
Complex Numbers and Quadratic Equation
117704
If \(z=r e^{i \theta}\), then \(\left|e^i\right|=\)
1 1
2 \(\mathrm{e}^{2 \mathrm{r} \sin \theta}\)
3 \(\mathrm{e}^{\mathrm{rsin} \theta}\)
4 \(\mathrm{e}^{-\mathrm{s} \sin \theta}\)
Explanation:
D Given, \(z=\mathrm{re}^{\mathrm{i} \theta}\) \(z=r(\cos \theta+i \sin \theta)\) \{From euller's formula \} Multiply by \(i\) to both side - \(\mathrm{iz}=\mathrm{ir}(\cos \theta+\mathrm{i} \sin \theta)\) \(\mathrm{iz}=\mathrm{ir} \cos \theta+\mathrm{i}^2 \mathrm{r} \sin \theta\) \(\mathrm{iz}=-\mathrm{r} \sin \theta+\mathrm{ir} \cos \theta\) \(\mathrm{e}^{\mathrm{iz}}=\mathrm{e}^{(-\mathrm{r} \sin \theta+\mathrm{ir} \cos \theta)}\) Taking modulus of both side \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right| .\left|\mathrm{e}^{\mathrm{ir} \cos \theta}\right|\) \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right| \times 1\left\{\because\left|\mathrm{e}^{\mathrm{i} \theta}\right|=1\right\}\) \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right|\)
COMEDK-2017
Complex Numbers and Quadratic Equation
117705
If the complex number \(z\) lies on a circle with centre at the origin and radius \(\frac{1}{4}\), then the complex number \(-1+8 \mathrm{z}\) lies on a circle with radius
1 4
2 1
3 3
4 2
Explanation:
D Given, \(\mathrm{z}\) is complex number lies on a circle with radius \(\frac{1}{4}\) we know that, \(|z|=r\) \(\therefore \quad|z| =\frac{1}{4}\) Let \(z^{\prime}=-1+8 z\) \(z=\frac{z^{\prime}+1}{8}\) Taking modules of both side \(|z|=\left|\frac{z^{\prime}+1}{8}\right|\) \(\frac{1}{4}=\left|\frac{z^{\prime}+1}{8}\right|\) \(\left|z^{\prime}+1\right|=2\) So, complex number \(z^{\prime}\) lies on circle with centre at \((-1\), \(0)\) and radius 2 .
VITEEE-2015
Complex Numbers and Quadratic Equation
117706
If the area of the triangle on the complex plane formed by the points \(z, z+i z\) and iz is 200 , then the value of \(3|z|\) must be equal to
1 20
2 40
3 60
4 80
Explanation:
C Let, \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(z+i z=x+i y+i(x+i y)\) \(=x-y+i x+i y \because i^2=-1\) \(=(x-y)+i(x+y)\) \(i z=i(x+i y) \because i^2=1\) \(=i x+i^2 y\) and \(\mathrm{iz} =\mathrm{i}(\mathrm{x}+\mathrm{iy})\) \(=\mathrm{ix}+\mathrm{i}^2 \mathrm{y} \because \mathrm{i}^2=1\) \(=-\mathrm{y}+\mathrm{ix}\) Then, the area of the triangle formed by these lines is \begin{align*} \begin{aligned} & \Delta=\frac{1}{2}\left\|\begin{array}{ccc} \mathrm{x} & \mathrm{y} & 1 \\ (\mathrm{x}-\mathrm{y}) & (\mathrm{x}+\mathrm{y}) & 1 \\ -\mathrm{y} & \mathrm{x} & 1 \end{array}\right\| \\ & \text { Applying } \mathrm{R}_2 \rightarrow \mathrm{R}_2-\left(\mathrm{R}_1+\mathrm{R}_3\right), \\ & \Delta=\frac{1}{2}\left\|\begin{array}{ccc} \mathrm{x} & \mathrm{y} & 1 \\ 0 & 0 & -1 \\ -\mathrm{y} & \mathrm{x} & 1 \end{array}\right\| \end{aligned} \end{align*} \(=\frac{1}{2}\left(\mathrm{x}^2+\mathrm{y}^2\right)\) \(\frac{1}{2}|z|^2=200\) \(|z|^2=400\) \(|z|=20\) \(\therefore 3|z|=3 \times 20\) \(3|z|=60\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Complex Numbers and Quadratic Equation
117702
The multiplicative inverse of \(\frac{3+4 i}{4-5 i}\) is
1 \(\left(\frac{-8}{25}, \frac{31}{25}\right)\)
2 \(\left(\frac{-8}{25}, \frac{-31}{25}\right)\)
3 \(\left(\frac{8}{25}, \frac{-31}{25}\right)\)
4 \(\left(\frac{8}{25}, \frac{31}{25}\right)\)
Explanation:
B Let \(z=\frac{3+4 i}{4-5 i}\) We have to calculate \(\mathrm{z}^{-1}\) i.e., \(\frac{1}{\mathrm{z}}\) \(\therefore z^{-1}=\frac{1}{z}=\frac{4-5 i}{3+4 i} \times \frac{3-4 i}{3-4 i}\) \(=\frac{12-15 i-16 i+20 i^2}{9-16 i^2}=\frac{12-31 i-20}{9-16(-1)}\) \(=\frac{-8-31 \mathrm{i}}{9+16}=\left(-\frac{8}{25}, \frac{-31}{25}\right)\)
COMEDK-2012
Complex Numbers and Quadratic Equation
117703
If a complex number lies in the III quadrant. Find the quadrant in which its conjugate lies.
1 I quadrant
2 II quadrant
3 III quadrant
4 IV quadrant
Explanation:
B Given, complex number lies in III quadrant. \(\therefore \quad \mathrm{z}=-\mathrm{a}-\mathrm{ib}\) \(\mathrm{a}\lt 0 \& \mathrm{~b}\lt 0\) So, it's conjugate is \(\bar{z}=-a+i b\) here \(\mathrm{a}\lt 0 \& \mathrm{~b}>0\) \(\therefore\) Conjugate lies in II quadrant.
COMEDK-2014
Complex Numbers and Quadratic Equation
117704
If \(z=r e^{i \theta}\), then \(\left|e^i\right|=\)
1 1
2 \(\mathrm{e}^{2 \mathrm{r} \sin \theta}\)
3 \(\mathrm{e}^{\mathrm{rsin} \theta}\)
4 \(\mathrm{e}^{-\mathrm{s} \sin \theta}\)
Explanation:
D Given, \(z=\mathrm{re}^{\mathrm{i} \theta}\) \(z=r(\cos \theta+i \sin \theta)\) \{From euller's formula \} Multiply by \(i\) to both side - \(\mathrm{iz}=\mathrm{ir}(\cos \theta+\mathrm{i} \sin \theta)\) \(\mathrm{iz}=\mathrm{ir} \cos \theta+\mathrm{i}^2 \mathrm{r} \sin \theta\) \(\mathrm{iz}=-\mathrm{r} \sin \theta+\mathrm{ir} \cos \theta\) \(\mathrm{e}^{\mathrm{iz}}=\mathrm{e}^{(-\mathrm{r} \sin \theta+\mathrm{ir} \cos \theta)}\) Taking modulus of both side \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right| .\left|\mathrm{e}^{\mathrm{ir} \cos \theta}\right|\) \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right| \times 1\left\{\because\left|\mathrm{e}^{\mathrm{i} \theta}\right|=1\right\}\) \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right|\)
COMEDK-2017
Complex Numbers and Quadratic Equation
117705
If the complex number \(z\) lies on a circle with centre at the origin and radius \(\frac{1}{4}\), then the complex number \(-1+8 \mathrm{z}\) lies on a circle with radius
1 4
2 1
3 3
4 2
Explanation:
D Given, \(\mathrm{z}\) is complex number lies on a circle with radius \(\frac{1}{4}\) we know that, \(|z|=r\) \(\therefore \quad|z| =\frac{1}{4}\) Let \(z^{\prime}=-1+8 z\) \(z=\frac{z^{\prime}+1}{8}\) Taking modules of both side \(|z|=\left|\frac{z^{\prime}+1}{8}\right|\) \(\frac{1}{4}=\left|\frac{z^{\prime}+1}{8}\right|\) \(\left|z^{\prime}+1\right|=2\) So, complex number \(z^{\prime}\) lies on circle with centre at \((-1\), \(0)\) and radius 2 .
VITEEE-2015
Complex Numbers and Quadratic Equation
117706
If the area of the triangle on the complex plane formed by the points \(z, z+i z\) and iz is 200 , then the value of \(3|z|\) must be equal to
1 20
2 40
3 60
4 80
Explanation:
C Let, \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(z+i z=x+i y+i(x+i y)\) \(=x-y+i x+i y \because i^2=-1\) \(=(x-y)+i(x+y)\) \(i z=i(x+i y) \because i^2=1\) \(=i x+i^2 y\) and \(\mathrm{iz} =\mathrm{i}(\mathrm{x}+\mathrm{iy})\) \(=\mathrm{ix}+\mathrm{i}^2 \mathrm{y} \because \mathrm{i}^2=1\) \(=-\mathrm{y}+\mathrm{ix}\) Then, the area of the triangle formed by these lines is \begin{align*} \begin{aligned} & \Delta=\frac{1}{2}\left\|\begin{array}{ccc} \mathrm{x} & \mathrm{y} & 1 \\ (\mathrm{x}-\mathrm{y}) & (\mathrm{x}+\mathrm{y}) & 1 \\ -\mathrm{y} & \mathrm{x} & 1 \end{array}\right\| \\ & \text { Applying } \mathrm{R}_2 \rightarrow \mathrm{R}_2-\left(\mathrm{R}_1+\mathrm{R}_3\right), \\ & \Delta=\frac{1}{2}\left\|\begin{array}{ccc} \mathrm{x} & \mathrm{y} & 1 \\ 0 & 0 & -1 \\ -\mathrm{y} & \mathrm{x} & 1 \end{array}\right\| \end{aligned} \end{align*} \(=\frac{1}{2}\left(\mathrm{x}^2+\mathrm{y}^2\right)\) \(\frac{1}{2}|z|^2=200\) \(|z|^2=400\) \(|z|=20\) \(\therefore 3|z|=3 \times 20\) \(3|z|=60\)
117702
The multiplicative inverse of \(\frac{3+4 i}{4-5 i}\) is
1 \(\left(\frac{-8}{25}, \frac{31}{25}\right)\)
2 \(\left(\frac{-8}{25}, \frac{-31}{25}\right)\)
3 \(\left(\frac{8}{25}, \frac{-31}{25}\right)\)
4 \(\left(\frac{8}{25}, \frac{31}{25}\right)\)
Explanation:
B Let \(z=\frac{3+4 i}{4-5 i}\) We have to calculate \(\mathrm{z}^{-1}\) i.e., \(\frac{1}{\mathrm{z}}\) \(\therefore z^{-1}=\frac{1}{z}=\frac{4-5 i}{3+4 i} \times \frac{3-4 i}{3-4 i}\) \(=\frac{12-15 i-16 i+20 i^2}{9-16 i^2}=\frac{12-31 i-20}{9-16(-1)}\) \(=\frac{-8-31 \mathrm{i}}{9+16}=\left(-\frac{8}{25}, \frac{-31}{25}\right)\)
COMEDK-2012
Complex Numbers and Quadratic Equation
117703
If a complex number lies in the III quadrant. Find the quadrant in which its conjugate lies.
1 I quadrant
2 II quadrant
3 III quadrant
4 IV quadrant
Explanation:
B Given, complex number lies in III quadrant. \(\therefore \quad \mathrm{z}=-\mathrm{a}-\mathrm{ib}\) \(\mathrm{a}\lt 0 \& \mathrm{~b}\lt 0\) So, it's conjugate is \(\bar{z}=-a+i b\) here \(\mathrm{a}\lt 0 \& \mathrm{~b}>0\) \(\therefore\) Conjugate lies in II quadrant.
COMEDK-2014
Complex Numbers and Quadratic Equation
117704
If \(z=r e^{i \theta}\), then \(\left|e^i\right|=\)
1 1
2 \(\mathrm{e}^{2 \mathrm{r} \sin \theta}\)
3 \(\mathrm{e}^{\mathrm{rsin} \theta}\)
4 \(\mathrm{e}^{-\mathrm{s} \sin \theta}\)
Explanation:
D Given, \(z=\mathrm{re}^{\mathrm{i} \theta}\) \(z=r(\cos \theta+i \sin \theta)\) \{From euller's formula \} Multiply by \(i\) to both side - \(\mathrm{iz}=\mathrm{ir}(\cos \theta+\mathrm{i} \sin \theta)\) \(\mathrm{iz}=\mathrm{ir} \cos \theta+\mathrm{i}^2 \mathrm{r} \sin \theta\) \(\mathrm{iz}=-\mathrm{r} \sin \theta+\mathrm{ir} \cos \theta\) \(\mathrm{e}^{\mathrm{iz}}=\mathrm{e}^{(-\mathrm{r} \sin \theta+\mathrm{ir} \cos \theta)}\) Taking modulus of both side \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right| .\left|\mathrm{e}^{\mathrm{ir} \cos \theta}\right|\) \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right| \times 1\left\{\because\left|\mathrm{e}^{\mathrm{i} \theta}\right|=1\right\}\) \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right|\)
COMEDK-2017
Complex Numbers and Quadratic Equation
117705
If the complex number \(z\) lies on a circle with centre at the origin and radius \(\frac{1}{4}\), then the complex number \(-1+8 \mathrm{z}\) lies on a circle with radius
1 4
2 1
3 3
4 2
Explanation:
D Given, \(\mathrm{z}\) is complex number lies on a circle with radius \(\frac{1}{4}\) we know that, \(|z|=r\) \(\therefore \quad|z| =\frac{1}{4}\) Let \(z^{\prime}=-1+8 z\) \(z=\frac{z^{\prime}+1}{8}\) Taking modules of both side \(|z|=\left|\frac{z^{\prime}+1}{8}\right|\) \(\frac{1}{4}=\left|\frac{z^{\prime}+1}{8}\right|\) \(\left|z^{\prime}+1\right|=2\) So, complex number \(z^{\prime}\) lies on circle with centre at \((-1\), \(0)\) and radius 2 .
VITEEE-2015
Complex Numbers and Quadratic Equation
117706
If the area of the triangle on the complex plane formed by the points \(z, z+i z\) and iz is 200 , then the value of \(3|z|\) must be equal to
1 20
2 40
3 60
4 80
Explanation:
C Let, \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(z+i z=x+i y+i(x+i y)\) \(=x-y+i x+i y \because i^2=-1\) \(=(x-y)+i(x+y)\) \(i z=i(x+i y) \because i^2=1\) \(=i x+i^2 y\) and \(\mathrm{iz} =\mathrm{i}(\mathrm{x}+\mathrm{iy})\) \(=\mathrm{ix}+\mathrm{i}^2 \mathrm{y} \because \mathrm{i}^2=1\) \(=-\mathrm{y}+\mathrm{ix}\) Then, the area of the triangle formed by these lines is \begin{align*} \begin{aligned} & \Delta=\frac{1}{2}\left\|\begin{array}{ccc} \mathrm{x} & \mathrm{y} & 1 \\ (\mathrm{x}-\mathrm{y}) & (\mathrm{x}+\mathrm{y}) & 1 \\ -\mathrm{y} & \mathrm{x} & 1 \end{array}\right\| \\ & \text { Applying } \mathrm{R}_2 \rightarrow \mathrm{R}_2-\left(\mathrm{R}_1+\mathrm{R}_3\right), \\ & \Delta=\frac{1}{2}\left\|\begin{array}{ccc} \mathrm{x} & \mathrm{y} & 1 \\ 0 & 0 & -1 \\ -\mathrm{y} & \mathrm{x} & 1 \end{array}\right\| \end{aligned} \end{align*} \(=\frac{1}{2}\left(\mathrm{x}^2+\mathrm{y}^2\right)\) \(\frac{1}{2}|z|^2=200\) \(|z|^2=400\) \(|z|=20\) \(\therefore 3|z|=3 \times 20\) \(3|z|=60\)
117702
The multiplicative inverse of \(\frac{3+4 i}{4-5 i}\) is
1 \(\left(\frac{-8}{25}, \frac{31}{25}\right)\)
2 \(\left(\frac{-8}{25}, \frac{-31}{25}\right)\)
3 \(\left(\frac{8}{25}, \frac{-31}{25}\right)\)
4 \(\left(\frac{8}{25}, \frac{31}{25}\right)\)
Explanation:
B Let \(z=\frac{3+4 i}{4-5 i}\) We have to calculate \(\mathrm{z}^{-1}\) i.e., \(\frac{1}{\mathrm{z}}\) \(\therefore z^{-1}=\frac{1}{z}=\frac{4-5 i}{3+4 i} \times \frac{3-4 i}{3-4 i}\) \(=\frac{12-15 i-16 i+20 i^2}{9-16 i^2}=\frac{12-31 i-20}{9-16(-1)}\) \(=\frac{-8-31 \mathrm{i}}{9+16}=\left(-\frac{8}{25}, \frac{-31}{25}\right)\)
COMEDK-2012
Complex Numbers and Quadratic Equation
117703
If a complex number lies in the III quadrant. Find the quadrant in which its conjugate lies.
1 I quadrant
2 II quadrant
3 III quadrant
4 IV quadrant
Explanation:
B Given, complex number lies in III quadrant. \(\therefore \quad \mathrm{z}=-\mathrm{a}-\mathrm{ib}\) \(\mathrm{a}\lt 0 \& \mathrm{~b}\lt 0\) So, it's conjugate is \(\bar{z}=-a+i b\) here \(\mathrm{a}\lt 0 \& \mathrm{~b}>0\) \(\therefore\) Conjugate lies in II quadrant.
COMEDK-2014
Complex Numbers and Quadratic Equation
117704
If \(z=r e^{i \theta}\), then \(\left|e^i\right|=\)
1 1
2 \(\mathrm{e}^{2 \mathrm{r} \sin \theta}\)
3 \(\mathrm{e}^{\mathrm{rsin} \theta}\)
4 \(\mathrm{e}^{-\mathrm{s} \sin \theta}\)
Explanation:
D Given, \(z=\mathrm{re}^{\mathrm{i} \theta}\) \(z=r(\cos \theta+i \sin \theta)\) \{From euller's formula \} Multiply by \(i\) to both side - \(\mathrm{iz}=\mathrm{ir}(\cos \theta+\mathrm{i} \sin \theta)\) \(\mathrm{iz}=\mathrm{ir} \cos \theta+\mathrm{i}^2 \mathrm{r} \sin \theta\) \(\mathrm{iz}=-\mathrm{r} \sin \theta+\mathrm{ir} \cos \theta\) \(\mathrm{e}^{\mathrm{iz}}=\mathrm{e}^{(-\mathrm{r} \sin \theta+\mathrm{ir} \cos \theta)}\) Taking modulus of both side \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right| .\left|\mathrm{e}^{\mathrm{ir} \cos \theta}\right|\) \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right| \times 1\left\{\because\left|\mathrm{e}^{\mathrm{i} \theta}\right|=1\right\}\) \(\left|\mathrm{e}^{\mathrm{iz}}\right|=\left|\mathrm{e}^{-\mathrm{r} \sin \theta}\right|\)
COMEDK-2017
Complex Numbers and Quadratic Equation
117705
If the complex number \(z\) lies on a circle with centre at the origin and radius \(\frac{1}{4}\), then the complex number \(-1+8 \mathrm{z}\) lies on a circle with radius
1 4
2 1
3 3
4 2
Explanation:
D Given, \(\mathrm{z}\) is complex number lies on a circle with radius \(\frac{1}{4}\) we know that, \(|z|=r\) \(\therefore \quad|z| =\frac{1}{4}\) Let \(z^{\prime}=-1+8 z\) \(z=\frac{z^{\prime}+1}{8}\) Taking modules of both side \(|z|=\left|\frac{z^{\prime}+1}{8}\right|\) \(\frac{1}{4}=\left|\frac{z^{\prime}+1}{8}\right|\) \(\left|z^{\prime}+1\right|=2\) So, complex number \(z^{\prime}\) lies on circle with centre at \((-1\), \(0)\) and radius 2 .
VITEEE-2015
Complex Numbers and Quadratic Equation
117706
If the area of the triangle on the complex plane formed by the points \(z, z+i z\) and iz is 200 , then the value of \(3|z|\) must be equal to
1 20
2 40
3 60
4 80
Explanation:
C Let, \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(z+i z=x+i y+i(x+i y)\) \(=x-y+i x+i y \because i^2=-1\) \(=(x-y)+i(x+y)\) \(i z=i(x+i y) \because i^2=1\) \(=i x+i^2 y\) and \(\mathrm{iz} =\mathrm{i}(\mathrm{x}+\mathrm{iy})\) \(=\mathrm{ix}+\mathrm{i}^2 \mathrm{y} \because \mathrm{i}^2=1\) \(=-\mathrm{y}+\mathrm{ix}\) Then, the area of the triangle formed by these lines is \begin{align*} \begin{aligned} & \Delta=\frac{1}{2}\left\|\begin{array}{ccc} \mathrm{x} & \mathrm{y} & 1 \\ (\mathrm{x}-\mathrm{y}) & (\mathrm{x}+\mathrm{y}) & 1 \\ -\mathrm{y} & \mathrm{x} & 1 \end{array}\right\| \\ & \text { Applying } \mathrm{R}_2 \rightarrow \mathrm{R}_2-\left(\mathrm{R}_1+\mathrm{R}_3\right), \\ & \Delta=\frac{1}{2}\left\|\begin{array}{ccc} \mathrm{x} & \mathrm{y} & 1 \\ 0 & 0 & -1 \\ -\mathrm{y} & \mathrm{x} & 1 \end{array}\right\| \end{aligned} \end{align*} \(=\frac{1}{2}\left(\mathrm{x}^2+\mathrm{y}^2\right)\) \(\frac{1}{2}|z|^2=200\) \(|z|^2=400\) \(|z|=20\) \(\therefore 3|z|=3 \times 20\) \(3|z|=60\)