117508
If \(z\) satisfies the equation \(|z|-z=1+2 i\), then \(z\) is equal to
1 \(\frac{3}{2}+2 \mathrm{i}\)
2 \(\frac{3}{2}-2 \mathrm{i}\)
3 \(2-\frac{3}{2} \mathrm{i}\)
4 \(2+\frac{3}{2} \mathrm{i}\)
Explanation:
B Given : \(|z|-z=1+2 i\) If \(z=x+i y\), then this equation reduces to \(|\mathrm{x}+\mathrm{iy}|-(\mathrm{x}+\mathrm{iy})=1+2 \mathrm{i}\) \(\Rightarrow\left(\sqrt{\mathrm{x}^2+\mathrm{y}^2}-\mathrm{x}\right)+(-\mathrm{iy})=1+2 \mathrm{i}\) On comparing real and imaginary parts of both sides of the equation, we get \(\sqrt{x^2+y^2}-x=1\) \(\Rightarrow \sqrt{x^2+y^2}=1+x \Rightarrow x^2+y^2=(1+x)^2\) \(\Rightarrow x^2+y^2=1+x^2+2 x\) \(\Rightarrow y^2=1+2 x\) \(\text { and }-y=2\) \(\Rightarrow y=-2\) Putting this value in eq. (i), we get \((-2)^2=1+2 \mathrm{x}\) \(\Rightarrow 2 \mathrm{x}=3 \Rightarrow \mathrm{x}=\frac{3}{2}\) \(\therefore \mathrm{z}=\mathrm{x}+\mathrm{y}=\frac{3}{2}-2 \mathrm{i}\)
VITEEE-2010
Complex Numbers and Quadratic Equation
117509
The perimeter of the locus represented by arg \(\left(\frac{z+i}{z-i}\right)=\frac{\pi}{4}\) is equal to
1 \(4 \pi\)
2 \(2 \pi \sqrt{2}\)
3 \(2 \pi \sqrt{3}\)
4 \(\frac{2 \pi}{\sqrt{3}}\)
Explanation:
B Locus represented by \(\arg \left(\frac{z+i}{z-i}\right)=\frac{\pi}{4}\) \(=\quad \arg (z+i)-\arg (z-i)=\frac{\pi}{4}\) \(=\quad \arg (x+i y+i)-\arg (x+i y-i)=\frac{\pi}{4}\) \(\arg (\mathrm{x}+\mathrm{i}(\mathrm{y}+1)\}-\arg \{\mathrm{x}+\mathrm{i}(\mathrm{y}-1)\}=\frac{\pi}{4}\) \(\tan ^{-1} \frac{(\mathrm{y}+1)}{\mathrm{x}}-\tan ^{-1} \frac{(\mathrm{y}-1)}{\mathrm{x}}=\frac{\pi}{4}\) \(\tan ^{-1}\left\{\frac{\frac{y+1}{x}-\frac{(y-1)}{x}}{1+\frac{(y+1)}{x} \frac{(y-1)}{x}}\right\}=\frac{\pi}{4}\) \(\tan ^{-1}\left\{\frac{\frac{2}{x}}{1+\frac{\left(y^2-1\right)}{x^2}}\right\}=\frac{\pi}{4}\) \(\frac{2 \mathrm{x}}{\mathrm{x}^2+\mathrm{y}^2-1}=\tan \frac{\pi}{4}\) \(\frac{2 \mathrm{x}}{\mathrm{x}^2+\mathrm{y}^2-1}=1\) \(\mathrm{x}^2+\mathrm{y}^2-1=2 \mathrm{x}\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}=1\) \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2=1+1\) \((\mathrm{x}-1)^2+\mathrm{y}^2=2\) It's represent a locus of circle with radius \(\sqrt{2}\) So, perimeter is \(2 \pi \mathrm{r}\) \(=\quad 2 \pi \sqrt{2}\)
UPSEE-2018
Complex Numbers and Quadratic Equation
117510
If \(\cos \left(\log \mathrm{i}^{44}\right)=\mathbf{a}+\mathrm{ib}\), then
117511
If \(\mathrm{z}\) is a complex number, then \((\mathrm{z}+5)(\overline{\mathrm{z}}+5)\) is equal to
1 \(|z+5 i|^2\)
2 \(|z-5|^2\)
3 \((z+5)^2\)
4 \(|z+5|^2\)
Explanation:
D Given, \(\mathrm{z}\) is complex number then, \(= (z+5)(\bar{z}+5)\) \(= (z+5)(\overline{z+5})\) Now, we know that - \(\overline{-}=|z|^2\) \(\therefore \quad (z+5)(\overline{z+5})=|z+5|^2\)
UPSEE-2016
Complex Numbers and Quadratic Equation
117512
If \(z\) is a complex number, then which of the following statement is true?
1 ( \(z \bar{z})\) is purely imaginary
2 \((z \bar{z})\) is non-negative real
3 \((z-\bar{z})\) is purely real
4 \((z+\bar{z})\) is purely imaginary
Explanation:
B If \(z\) is a complex number Then, \(z=x+i y\) \(\bar{z}=x-i y\) Now, \(z \bar{z}=(x+i y)(x-i y)\) \(=x^2-(i y)^2\) \(=x^2-i^2 y^2\) \(z \bar{z}=x^2+y^2\left\{\because i^2=-1\right\}\) So, from above equation \(\mathrm{z} \overline{\mathrm{Z}}\) is non-negative real.
117508
If \(z\) satisfies the equation \(|z|-z=1+2 i\), then \(z\) is equal to
1 \(\frac{3}{2}+2 \mathrm{i}\)
2 \(\frac{3}{2}-2 \mathrm{i}\)
3 \(2-\frac{3}{2} \mathrm{i}\)
4 \(2+\frac{3}{2} \mathrm{i}\)
Explanation:
B Given : \(|z|-z=1+2 i\) If \(z=x+i y\), then this equation reduces to \(|\mathrm{x}+\mathrm{iy}|-(\mathrm{x}+\mathrm{iy})=1+2 \mathrm{i}\) \(\Rightarrow\left(\sqrt{\mathrm{x}^2+\mathrm{y}^2}-\mathrm{x}\right)+(-\mathrm{iy})=1+2 \mathrm{i}\) On comparing real and imaginary parts of both sides of the equation, we get \(\sqrt{x^2+y^2}-x=1\) \(\Rightarrow \sqrt{x^2+y^2}=1+x \Rightarrow x^2+y^2=(1+x)^2\) \(\Rightarrow x^2+y^2=1+x^2+2 x\) \(\Rightarrow y^2=1+2 x\) \(\text { and }-y=2\) \(\Rightarrow y=-2\) Putting this value in eq. (i), we get \((-2)^2=1+2 \mathrm{x}\) \(\Rightarrow 2 \mathrm{x}=3 \Rightarrow \mathrm{x}=\frac{3}{2}\) \(\therefore \mathrm{z}=\mathrm{x}+\mathrm{y}=\frac{3}{2}-2 \mathrm{i}\)
VITEEE-2010
Complex Numbers and Quadratic Equation
117509
The perimeter of the locus represented by arg \(\left(\frac{z+i}{z-i}\right)=\frac{\pi}{4}\) is equal to
1 \(4 \pi\)
2 \(2 \pi \sqrt{2}\)
3 \(2 \pi \sqrt{3}\)
4 \(\frac{2 \pi}{\sqrt{3}}\)
Explanation:
B Locus represented by \(\arg \left(\frac{z+i}{z-i}\right)=\frac{\pi}{4}\) \(=\quad \arg (z+i)-\arg (z-i)=\frac{\pi}{4}\) \(=\quad \arg (x+i y+i)-\arg (x+i y-i)=\frac{\pi}{4}\) \(\arg (\mathrm{x}+\mathrm{i}(\mathrm{y}+1)\}-\arg \{\mathrm{x}+\mathrm{i}(\mathrm{y}-1)\}=\frac{\pi}{4}\) \(\tan ^{-1} \frac{(\mathrm{y}+1)}{\mathrm{x}}-\tan ^{-1} \frac{(\mathrm{y}-1)}{\mathrm{x}}=\frac{\pi}{4}\) \(\tan ^{-1}\left\{\frac{\frac{y+1}{x}-\frac{(y-1)}{x}}{1+\frac{(y+1)}{x} \frac{(y-1)}{x}}\right\}=\frac{\pi}{4}\) \(\tan ^{-1}\left\{\frac{\frac{2}{x}}{1+\frac{\left(y^2-1\right)}{x^2}}\right\}=\frac{\pi}{4}\) \(\frac{2 \mathrm{x}}{\mathrm{x}^2+\mathrm{y}^2-1}=\tan \frac{\pi}{4}\) \(\frac{2 \mathrm{x}}{\mathrm{x}^2+\mathrm{y}^2-1}=1\) \(\mathrm{x}^2+\mathrm{y}^2-1=2 \mathrm{x}\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}=1\) \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2=1+1\) \((\mathrm{x}-1)^2+\mathrm{y}^2=2\) It's represent a locus of circle with radius \(\sqrt{2}\) So, perimeter is \(2 \pi \mathrm{r}\) \(=\quad 2 \pi \sqrt{2}\)
UPSEE-2018
Complex Numbers and Quadratic Equation
117510
If \(\cos \left(\log \mathrm{i}^{44}\right)=\mathbf{a}+\mathrm{ib}\), then
117511
If \(\mathrm{z}\) is a complex number, then \((\mathrm{z}+5)(\overline{\mathrm{z}}+5)\) is equal to
1 \(|z+5 i|^2\)
2 \(|z-5|^2\)
3 \((z+5)^2\)
4 \(|z+5|^2\)
Explanation:
D Given, \(\mathrm{z}\) is complex number then, \(= (z+5)(\bar{z}+5)\) \(= (z+5)(\overline{z+5})\) Now, we know that - \(\overline{-}=|z|^2\) \(\therefore \quad (z+5)(\overline{z+5})=|z+5|^2\)
UPSEE-2016
Complex Numbers and Quadratic Equation
117512
If \(z\) is a complex number, then which of the following statement is true?
1 ( \(z \bar{z})\) is purely imaginary
2 \((z \bar{z})\) is non-negative real
3 \((z-\bar{z})\) is purely real
4 \((z+\bar{z})\) is purely imaginary
Explanation:
B If \(z\) is a complex number Then, \(z=x+i y\) \(\bar{z}=x-i y\) Now, \(z \bar{z}=(x+i y)(x-i y)\) \(=x^2-(i y)^2\) \(=x^2-i^2 y^2\) \(z \bar{z}=x^2+y^2\left\{\because i^2=-1\right\}\) So, from above equation \(\mathrm{z} \overline{\mathrm{Z}}\) is non-negative real.
117508
If \(z\) satisfies the equation \(|z|-z=1+2 i\), then \(z\) is equal to
1 \(\frac{3}{2}+2 \mathrm{i}\)
2 \(\frac{3}{2}-2 \mathrm{i}\)
3 \(2-\frac{3}{2} \mathrm{i}\)
4 \(2+\frac{3}{2} \mathrm{i}\)
Explanation:
B Given : \(|z|-z=1+2 i\) If \(z=x+i y\), then this equation reduces to \(|\mathrm{x}+\mathrm{iy}|-(\mathrm{x}+\mathrm{iy})=1+2 \mathrm{i}\) \(\Rightarrow\left(\sqrt{\mathrm{x}^2+\mathrm{y}^2}-\mathrm{x}\right)+(-\mathrm{iy})=1+2 \mathrm{i}\) On comparing real and imaginary parts of both sides of the equation, we get \(\sqrt{x^2+y^2}-x=1\) \(\Rightarrow \sqrt{x^2+y^2}=1+x \Rightarrow x^2+y^2=(1+x)^2\) \(\Rightarrow x^2+y^2=1+x^2+2 x\) \(\Rightarrow y^2=1+2 x\) \(\text { and }-y=2\) \(\Rightarrow y=-2\) Putting this value in eq. (i), we get \((-2)^2=1+2 \mathrm{x}\) \(\Rightarrow 2 \mathrm{x}=3 \Rightarrow \mathrm{x}=\frac{3}{2}\) \(\therefore \mathrm{z}=\mathrm{x}+\mathrm{y}=\frac{3}{2}-2 \mathrm{i}\)
VITEEE-2010
Complex Numbers and Quadratic Equation
117509
The perimeter of the locus represented by arg \(\left(\frac{z+i}{z-i}\right)=\frac{\pi}{4}\) is equal to
1 \(4 \pi\)
2 \(2 \pi \sqrt{2}\)
3 \(2 \pi \sqrt{3}\)
4 \(\frac{2 \pi}{\sqrt{3}}\)
Explanation:
B Locus represented by \(\arg \left(\frac{z+i}{z-i}\right)=\frac{\pi}{4}\) \(=\quad \arg (z+i)-\arg (z-i)=\frac{\pi}{4}\) \(=\quad \arg (x+i y+i)-\arg (x+i y-i)=\frac{\pi}{4}\) \(\arg (\mathrm{x}+\mathrm{i}(\mathrm{y}+1)\}-\arg \{\mathrm{x}+\mathrm{i}(\mathrm{y}-1)\}=\frac{\pi}{4}\) \(\tan ^{-1} \frac{(\mathrm{y}+1)}{\mathrm{x}}-\tan ^{-1} \frac{(\mathrm{y}-1)}{\mathrm{x}}=\frac{\pi}{4}\) \(\tan ^{-1}\left\{\frac{\frac{y+1}{x}-\frac{(y-1)}{x}}{1+\frac{(y+1)}{x} \frac{(y-1)}{x}}\right\}=\frac{\pi}{4}\) \(\tan ^{-1}\left\{\frac{\frac{2}{x}}{1+\frac{\left(y^2-1\right)}{x^2}}\right\}=\frac{\pi}{4}\) \(\frac{2 \mathrm{x}}{\mathrm{x}^2+\mathrm{y}^2-1}=\tan \frac{\pi}{4}\) \(\frac{2 \mathrm{x}}{\mathrm{x}^2+\mathrm{y}^2-1}=1\) \(\mathrm{x}^2+\mathrm{y}^2-1=2 \mathrm{x}\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}=1\) \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2=1+1\) \((\mathrm{x}-1)^2+\mathrm{y}^2=2\) It's represent a locus of circle with radius \(\sqrt{2}\) So, perimeter is \(2 \pi \mathrm{r}\) \(=\quad 2 \pi \sqrt{2}\)
UPSEE-2018
Complex Numbers and Quadratic Equation
117510
If \(\cos \left(\log \mathrm{i}^{44}\right)=\mathbf{a}+\mathrm{ib}\), then
117511
If \(\mathrm{z}\) is a complex number, then \((\mathrm{z}+5)(\overline{\mathrm{z}}+5)\) is equal to
1 \(|z+5 i|^2\)
2 \(|z-5|^2\)
3 \((z+5)^2\)
4 \(|z+5|^2\)
Explanation:
D Given, \(\mathrm{z}\) is complex number then, \(= (z+5)(\bar{z}+5)\) \(= (z+5)(\overline{z+5})\) Now, we know that - \(\overline{-}=|z|^2\) \(\therefore \quad (z+5)(\overline{z+5})=|z+5|^2\)
UPSEE-2016
Complex Numbers and Quadratic Equation
117512
If \(z\) is a complex number, then which of the following statement is true?
1 ( \(z \bar{z})\) is purely imaginary
2 \((z \bar{z})\) is non-negative real
3 \((z-\bar{z})\) is purely real
4 \((z+\bar{z})\) is purely imaginary
Explanation:
B If \(z\) is a complex number Then, \(z=x+i y\) \(\bar{z}=x-i y\) Now, \(z \bar{z}=(x+i y)(x-i y)\) \(=x^2-(i y)^2\) \(=x^2-i^2 y^2\) \(z \bar{z}=x^2+y^2\left\{\because i^2=-1\right\}\) So, from above equation \(\mathrm{z} \overline{\mathrm{Z}}\) is non-negative real.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
117508
If \(z\) satisfies the equation \(|z|-z=1+2 i\), then \(z\) is equal to
1 \(\frac{3}{2}+2 \mathrm{i}\)
2 \(\frac{3}{2}-2 \mathrm{i}\)
3 \(2-\frac{3}{2} \mathrm{i}\)
4 \(2+\frac{3}{2} \mathrm{i}\)
Explanation:
B Given : \(|z|-z=1+2 i\) If \(z=x+i y\), then this equation reduces to \(|\mathrm{x}+\mathrm{iy}|-(\mathrm{x}+\mathrm{iy})=1+2 \mathrm{i}\) \(\Rightarrow\left(\sqrt{\mathrm{x}^2+\mathrm{y}^2}-\mathrm{x}\right)+(-\mathrm{iy})=1+2 \mathrm{i}\) On comparing real and imaginary parts of both sides of the equation, we get \(\sqrt{x^2+y^2}-x=1\) \(\Rightarrow \sqrt{x^2+y^2}=1+x \Rightarrow x^2+y^2=(1+x)^2\) \(\Rightarrow x^2+y^2=1+x^2+2 x\) \(\Rightarrow y^2=1+2 x\) \(\text { and }-y=2\) \(\Rightarrow y=-2\) Putting this value in eq. (i), we get \((-2)^2=1+2 \mathrm{x}\) \(\Rightarrow 2 \mathrm{x}=3 \Rightarrow \mathrm{x}=\frac{3}{2}\) \(\therefore \mathrm{z}=\mathrm{x}+\mathrm{y}=\frac{3}{2}-2 \mathrm{i}\)
VITEEE-2010
Complex Numbers and Quadratic Equation
117509
The perimeter of the locus represented by arg \(\left(\frac{z+i}{z-i}\right)=\frac{\pi}{4}\) is equal to
1 \(4 \pi\)
2 \(2 \pi \sqrt{2}\)
3 \(2 \pi \sqrt{3}\)
4 \(\frac{2 \pi}{\sqrt{3}}\)
Explanation:
B Locus represented by \(\arg \left(\frac{z+i}{z-i}\right)=\frac{\pi}{4}\) \(=\quad \arg (z+i)-\arg (z-i)=\frac{\pi}{4}\) \(=\quad \arg (x+i y+i)-\arg (x+i y-i)=\frac{\pi}{4}\) \(\arg (\mathrm{x}+\mathrm{i}(\mathrm{y}+1)\}-\arg \{\mathrm{x}+\mathrm{i}(\mathrm{y}-1)\}=\frac{\pi}{4}\) \(\tan ^{-1} \frac{(\mathrm{y}+1)}{\mathrm{x}}-\tan ^{-1} \frac{(\mathrm{y}-1)}{\mathrm{x}}=\frac{\pi}{4}\) \(\tan ^{-1}\left\{\frac{\frac{y+1}{x}-\frac{(y-1)}{x}}{1+\frac{(y+1)}{x} \frac{(y-1)}{x}}\right\}=\frac{\pi}{4}\) \(\tan ^{-1}\left\{\frac{\frac{2}{x}}{1+\frac{\left(y^2-1\right)}{x^2}}\right\}=\frac{\pi}{4}\) \(\frac{2 \mathrm{x}}{\mathrm{x}^2+\mathrm{y}^2-1}=\tan \frac{\pi}{4}\) \(\frac{2 \mathrm{x}}{\mathrm{x}^2+\mathrm{y}^2-1}=1\) \(\mathrm{x}^2+\mathrm{y}^2-1=2 \mathrm{x}\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}=1\) \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2=1+1\) \((\mathrm{x}-1)^2+\mathrm{y}^2=2\) It's represent a locus of circle with radius \(\sqrt{2}\) So, perimeter is \(2 \pi \mathrm{r}\) \(=\quad 2 \pi \sqrt{2}\)
UPSEE-2018
Complex Numbers and Quadratic Equation
117510
If \(\cos \left(\log \mathrm{i}^{44}\right)=\mathbf{a}+\mathrm{ib}\), then
117511
If \(\mathrm{z}\) is a complex number, then \((\mathrm{z}+5)(\overline{\mathrm{z}}+5)\) is equal to
1 \(|z+5 i|^2\)
2 \(|z-5|^2\)
3 \((z+5)^2\)
4 \(|z+5|^2\)
Explanation:
D Given, \(\mathrm{z}\) is complex number then, \(= (z+5)(\bar{z}+5)\) \(= (z+5)(\overline{z+5})\) Now, we know that - \(\overline{-}=|z|^2\) \(\therefore \quad (z+5)(\overline{z+5})=|z+5|^2\)
UPSEE-2016
Complex Numbers and Quadratic Equation
117512
If \(z\) is a complex number, then which of the following statement is true?
1 ( \(z \bar{z})\) is purely imaginary
2 \((z \bar{z})\) is non-negative real
3 \((z-\bar{z})\) is purely real
4 \((z+\bar{z})\) is purely imaginary
Explanation:
B If \(z\) is a complex number Then, \(z=x+i y\) \(\bar{z}=x-i y\) Now, \(z \bar{z}=(x+i y)(x-i y)\) \(=x^2-(i y)^2\) \(=x^2-i^2 y^2\) \(z \bar{z}=x^2+y^2\left\{\because i^2=-1\right\}\) So, from above equation \(\mathrm{z} \overline{\mathrm{Z}}\) is non-negative real.
117508
If \(z\) satisfies the equation \(|z|-z=1+2 i\), then \(z\) is equal to
1 \(\frac{3}{2}+2 \mathrm{i}\)
2 \(\frac{3}{2}-2 \mathrm{i}\)
3 \(2-\frac{3}{2} \mathrm{i}\)
4 \(2+\frac{3}{2} \mathrm{i}\)
Explanation:
B Given : \(|z|-z=1+2 i\) If \(z=x+i y\), then this equation reduces to \(|\mathrm{x}+\mathrm{iy}|-(\mathrm{x}+\mathrm{iy})=1+2 \mathrm{i}\) \(\Rightarrow\left(\sqrt{\mathrm{x}^2+\mathrm{y}^2}-\mathrm{x}\right)+(-\mathrm{iy})=1+2 \mathrm{i}\) On comparing real and imaginary parts of both sides of the equation, we get \(\sqrt{x^2+y^2}-x=1\) \(\Rightarrow \sqrt{x^2+y^2}=1+x \Rightarrow x^2+y^2=(1+x)^2\) \(\Rightarrow x^2+y^2=1+x^2+2 x\) \(\Rightarrow y^2=1+2 x\) \(\text { and }-y=2\) \(\Rightarrow y=-2\) Putting this value in eq. (i), we get \((-2)^2=1+2 \mathrm{x}\) \(\Rightarrow 2 \mathrm{x}=3 \Rightarrow \mathrm{x}=\frac{3}{2}\) \(\therefore \mathrm{z}=\mathrm{x}+\mathrm{y}=\frac{3}{2}-2 \mathrm{i}\)
VITEEE-2010
Complex Numbers and Quadratic Equation
117509
The perimeter of the locus represented by arg \(\left(\frac{z+i}{z-i}\right)=\frac{\pi}{4}\) is equal to
1 \(4 \pi\)
2 \(2 \pi \sqrt{2}\)
3 \(2 \pi \sqrt{3}\)
4 \(\frac{2 \pi}{\sqrt{3}}\)
Explanation:
B Locus represented by \(\arg \left(\frac{z+i}{z-i}\right)=\frac{\pi}{4}\) \(=\quad \arg (z+i)-\arg (z-i)=\frac{\pi}{4}\) \(=\quad \arg (x+i y+i)-\arg (x+i y-i)=\frac{\pi}{4}\) \(\arg (\mathrm{x}+\mathrm{i}(\mathrm{y}+1)\}-\arg \{\mathrm{x}+\mathrm{i}(\mathrm{y}-1)\}=\frac{\pi}{4}\) \(\tan ^{-1} \frac{(\mathrm{y}+1)}{\mathrm{x}}-\tan ^{-1} \frac{(\mathrm{y}-1)}{\mathrm{x}}=\frac{\pi}{4}\) \(\tan ^{-1}\left\{\frac{\frac{y+1}{x}-\frac{(y-1)}{x}}{1+\frac{(y+1)}{x} \frac{(y-1)}{x}}\right\}=\frac{\pi}{4}\) \(\tan ^{-1}\left\{\frac{\frac{2}{x}}{1+\frac{\left(y^2-1\right)}{x^2}}\right\}=\frac{\pi}{4}\) \(\frac{2 \mathrm{x}}{\mathrm{x}^2+\mathrm{y}^2-1}=\tan \frac{\pi}{4}\) \(\frac{2 \mathrm{x}}{\mathrm{x}^2+\mathrm{y}^2-1}=1\) \(\mathrm{x}^2+\mathrm{y}^2-1=2 \mathrm{x}\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}=1\) \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2=1+1\) \((\mathrm{x}-1)^2+\mathrm{y}^2=2\) It's represent a locus of circle with radius \(\sqrt{2}\) So, perimeter is \(2 \pi \mathrm{r}\) \(=\quad 2 \pi \sqrt{2}\)
UPSEE-2018
Complex Numbers and Quadratic Equation
117510
If \(\cos \left(\log \mathrm{i}^{44}\right)=\mathbf{a}+\mathrm{ib}\), then
117511
If \(\mathrm{z}\) is a complex number, then \((\mathrm{z}+5)(\overline{\mathrm{z}}+5)\) is equal to
1 \(|z+5 i|^2\)
2 \(|z-5|^2\)
3 \((z+5)^2\)
4 \(|z+5|^2\)
Explanation:
D Given, \(\mathrm{z}\) is complex number then, \(= (z+5)(\bar{z}+5)\) \(= (z+5)(\overline{z+5})\) Now, we know that - \(\overline{-}=|z|^2\) \(\therefore \quad (z+5)(\overline{z+5})=|z+5|^2\)
UPSEE-2016
Complex Numbers and Quadratic Equation
117512
If \(z\) is a complex number, then which of the following statement is true?
1 ( \(z \bar{z})\) is purely imaginary
2 \((z \bar{z})\) is non-negative real
3 \((z-\bar{z})\) is purely real
4 \((z+\bar{z})\) is purely imaginary
Explanation:
B If \(z\) is a complex number Then, \(z=x+i y\) \(\bar{z}=x-i y\) Now, \(z \bar{z}=(x+i y)(x-i y)\) \(=x^2-(i y)^2\) \(=x^2-i^2 y^2\) \(z \bar{z}=x^2+y^2\left\{\because i^2=-1\right\}\) So, from above equation \(\mathrm{z} \overline{\mathrm{Z}}\) is non-negative real.