A Let \(x=(16)^{1 / 4}\) \(x^4=16\) \(\mathrm{x}^4-16=0\) \(\left(\mathrm{x}^2\right)^2-\left(2^2\right)^2=0\) \(\left(x^2-2^2\right)\left(x^2+2^2\right)=0\) \((\mathrm{x}-2)(\mathrm{x}+2)\left(\mathrm{x}^2+4\right)=0\) \(\mathrm{x}= \pm 2\) \(x^2=-4\) \(x= \pm 2 \mathrm{i}\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117469
If \((x-i y)^{1 / 3}=a-i b\), then \(\frac{x}{a}+\frac{y}{b}\) is equal to
1 \(-2\left(a^2+b^2\right)\)
2 \(4(a+b)\)
3 \(4(a-b)\)
4 \(4 \mathrm{ab}\)
Explanation:
A Given, \((x-i y)^{1 / 3}=a-i b\) Now cubing both side \(\left\{(x-i y)^{1 / 3}\right\}^3=(a-i b)^3\) \((x-i y)=a^3-(i b)^3-3 a i b(a-i b)\) \(=a^3+i b^3-3 a^2 i b+\left(3 a i^2 b^2\right)\) \(=a^3+i b^3-3 a^2 b i-3 a b^2\) \(=\left(a^3-3 a b^2\right)+i\left(b^3-3 a^2 b\right)\) Comparing the value of \(x\) and \(y\) \(x=a\left(a^2-3 b^2\right) \text { and } y=b\left(b^2-3 a^2\right)\) Then, \(\frac{x}{a}+\frac{y}{b} =\frac{a\left(a^2-3 b^2\right)}{a}+\frac{b\left(b^2-3 a^2\right)}{b}\) \(\frac{x}{a}+\frac{y}{b} =a^2-3 b^2+b^2-3 a^2\) \(=-2 a^2-2 b^2\) \(\frac{x}{a}+\frac{y}{b} =-2\left(a^2+b^2\right)\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117470
If \((a+i b)(c+i d)(e+i f)(g+i h)=A+i B\), then \(\left(a^2+b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right)=\)
1 \(\mathrm{A}^2+\mathrm{B}^2\)
2 \(\mathrm{A}^2-\mathrm{B}^2\)
3 \(\mathrm{A}^2\)
4 \(\mathrm{B}^2\)
Explanation:
A Given, \((a+i b)(c+i d)(e+i f)(g+i h)=A+i B\) Then conjugate of \(\mathrm{A}+\mathrm{iB}\) is \(\mathrm{A}-\mathrm{iB}\) \(\Rightarrow \quad(\mathrm{a}-\mathrm{ib})(\mathrm{c}-\mathrm{id})(\mathrm{e}-\mathrm{if})(\mathrm{g}-\mathrm{ih})=\mathrm{A}-\mathrm{iB}\) Multiplying (1) and (2), we get \(\left(\mathrm{a}^2+\mathrm{b}^2\right)\left(\mathrm{c}^2+\mathrm{d}^2\right)\left(\mathrm{e}^2+\mathrm{f}^2\right)\left(\mathrm{g}^2+\mathrm{h}^2\right)=\mathrm{A}^2+\mathrm{B}^2\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117471
\(\quad(z+a)(\bar{z}+a)\), where \(a\) is real, is equivalent to
1 \(|z-a|\)
2 \(z^2+a^2\)
3 \(|z+a|^2\)
4 None of these
Explanation:
C \((z+a)(\bar{z}+a)=\text { ? }\) Given that \(a\) is real then, \(\overline{\mathrm{a}}=\mathrm{a}\) \((z+a)(\bar{z}+a)=(z+a)(\overline{z+a})\) We know that, \(z \bar{z}=|z|^2\)So, \(\quad(z+a)(\overline{z+a})=|z+a|^2\)
A Let \(x=(16)^{1 / 4}\) \(x^4=16\) \(\mathrm{x}^4-16=0\) \(\left(\mathrm{x}^2\right)^2-\left(2^2\right)^2=0\) \(\left(x^2-2^2\right)\left(x^2+2^2\right)=0\) \((\mathrm{x}-2)(\mathrm{x}+2)\left(\mathrm{x}^2+4\right)=0\) \(\mathrm{x}= \pm 2\) \(x^2=-4\) \(x= \pm 2 \mathrm{i}\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117469
If \((x-i y)^{1 / 3}=a-i b\), then \(\frac{x}{a}+\frac{y}{b}\) is equal to
1 \(-2\left(a^2+b^2\right)\)
2 \(4(a+b)\)
3 \(4(a-b)\)
4 \(4 \mathrm{ab}\)
Explanation:
A Given, \((x-i y)^{1 / 3}=a-i b\) Now cubing both side \(\left\{(x-i y)^{1 / 3}\right\}^3=(a-i b)^3\) \((x-i y)=a^3-(i b)^3-3 a i b(a-i b)\) \(=a^3+i b^3-3 a^2 i b+\left(3 a i^2 b^2\right)\) \(=a^3+i b^3-3 a^2 b i-3 a b^2\) \(=\left(a^3-3 a b^2\right)+i\left(b^3-3 a^2 b\right)\) Comparing the value of \(x\) and \(y\) \(x=a\left(a^2-3 b^2\right) \text { and } y=b\left(b^2-3 a^2\right)\) Then, \(\frac{x}{a}+\frac{y}{b} =\frac{a\left(a^2-3 b^2\right)}{a}+\frac{b\left(b^2-3 a^2\right)}{b}\) \(\frac{x}{a}+\frac{y}{b} =a^2-3 b^2+b^2-3 a^2\) \(=-2 a^2-2 b^2\) \(\frac{x}{a}+\frac{y}{b} =-2\left(a^2+b^2\right)\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117470
If \((a+i b)(c+i d)(e+i f)(g+i h)=A+i B\), then \(\left(a^2+b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right)=\)
1 \(\mathrm{A}^2+\mathrm{B}^2\)
2 \(\mathrm{A}^2-\mathrm{B}^2\)
3 \(\mathrm{A}^2\)
4 \(\mathrm{B}^2\)
Explanation:
A Given, \((a+i b)(c+i d)(e+i f)(g+i h)=A+i B\) Then conjugate of \(\mathrm{A}+\mathrm{iB}\) is \(\mathrm{A}-\mathrm{iB}\) \(\Rightarrow \quad(\mathrm{a}-\mathrm{ib})(\mathrm{c}-\mathrm{id})(\mathrm{e}-\mathrm{if})(\mathrm{g}-\mathrm{ih})=\mathrm{A}-\mathrm{iB}\) Multiplying (1) and (2), we get \(\left(\mathrm{a}^2+\mathrm{b}^2\right)\left(\mathrm{c}^2+\mathrm{d}^2\right)\left(\mathrm{e}^2+\mathrm{f}^2\right)\left(\mathrm{g}^2+\mathrm{h}^2\right)=\mathrm{A}^2+\mathrm{B}^2\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117471
\(\quad(z+a)(\bar{z}+a)\), where \(a\) is real, is equivalent to
1 \(|z-a|\)
2 \(z^2+a^2\)
3 \(|z+a|^2\)
4 None of these
Explanation:
C \((z+a)(\bar{z}+a)=\text { ? }\) Given that \(a\) is real then, \(\overline{\mathrm{a}}=\mathrm{a}\) \((z+a)(\bar{z}+a)=(z+a)(\overline{z+a})\) We know that, \(z \bar{z}=|z|^2\)So, \(\quad(z+a)(\overline{z+a})=|z+a|^2\)
A Let \(x=(16)^{1 / 4}\) \(x^4=16\) \(\mathrm{x}^4-16=0\) \(\left(\mathrm{x}^2\right)^2-\left(2^2\right)^2=0\) \(\left(x^2-2^2\right)\left(x^2+2^2\right)=0\) \((\mathrm{x}-2)(\mathrm{x}+2)\left(\mathrm{x}^2+4\right)=0\) \(\mathrm{x}= \pm 2\) \(x^2=-4\) \(x= \pm 2 \mathrm{i}\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117469
If \((x-i y)^{1 / 3}=a-i b\), then \(\frac{x}{a}+\frac{y}{b}\) is equal to
1 \(-2\left(a^2+b^2\right)\)
2 \(4(a+b)\)
3 \(4(a-b)\)
4 \(4 \mathrm{ab}\)
Explanation:
A Given, \((x-i y)^{1 / 3}=a-i b\) Now cubing both side \(\left\{(x-i y)^{1 / 3}\right\}^3=(a-i b)^3\) \((x-i y)=a^3-(i b)^3-3 a i b(a-i b)\) \(=a^3+i b^3-3 a^2 i b+\left(3 a i^2 b^2\right)\) \(=a^3+i b^3-3 a^2 b i-3 a b^2\) \(=\left(a^3-3 a b^2\right)+i\left(b^3-3 a^2 b\right)\) Comparing the value of \(x\) and \(y\) \(x=a\left(a^2-3 b^2\right) \text { and } y=b\left(b^2-3 a^2\right)\) Then, \(\frac{x}{a}+\frac{y}{b} =\frac{a\left(a^2-3 b^2\right)}{a}+\frac{b\left(b^2-3 a^2\right)}{b}\) \(\frac{x}{a}+\frac{y}{b} =a^2-3 b^2+b^2-3 a^2\) \(=-2 a^2-2 b^2\) \(\frac{x}{a}+\frac{y}{b} =-2\left(a^2+b^2\right)\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117470
If \((a+i b)(c+i d)(e+i f)(g+i h)=A+i B\), then \(\left(a^2+b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right)=\)
1 \(\mathrm{A}^2+\mathrm{B}^2\)
2 \(\mathrm{A}^2-\mathrm{B}^2\)
3 \(\mathrm{A}^2\)
4 \(\mathrm{B}^2\)
Explanation:
A Given, \((a+i b)(c+i d)(e+i f)(g+i h)=A+i B\) Then conjugate of \(\mathrm{A}+\mathrm{iB}\) is \(\mathrm{A}-\mathrm{iB}\) \(\Rightarrow \quad(\mathrm{a}-\mathrm{ib})(\mathrm{c}-\mathrm{id})(\mathrm{e}-\mathrm{if})(\mathrm{g}-\mathrm{ih})=\mathrm{A}-\mathrm{iB}\) Multiplying (1) and (2), we get \(\left(\mathrm{a}^2+\mathrm{b}^2\right)\left(\mathrm{c}^2+\mathrm{d}^2\right)\left(\mathrm{e}^2+\mathrm{f}^2\right)\left(\mathrm{g}^2+\mathrm{h}^2\right)=\mathrm{A}^2+\mathrm{B}^2\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117471
\(\quad(z+a)(\bar{z}+a)\), where \(a\) is real, is equivalent to
1 \(|z-a|\)
2 \(z^2+a^2\)
3 \(|z+a|^2\)
4 None of these
Explanation:
C \((z+a)(\bar{z}+a)=\text { ? }\) Given that \(a\) is real then, \(\overline{\mathrm{a}}=\mathrm{a}\) \((z+a)(\bar{z}+a)=(z+a)(\overline{z+a})\) We know that, \(z \bar{z}=|z|^2\)So, \(\quad(z+a)(\overline{z+a})=|z+a|^2\)
A Let \(x=(16)^{1 / 4}\) \(x^4=16\) \(\mathrm{x}^4-16=0\) \(\left(\mathrm{x}^2\right)^2-\left(2^2\right)^2=0\) \(\left(x^2-2^2\right)\left(x^2+2^2\right)=0\) \((\mathrm{x}-2)(\mathrm{x}+2)\left(\mathrm{x}^2+4\right)=0\) \(\mathrm{x}= \pm 2\) \(x^2=-4\) \(x= \pm 2 \mathrm{i}\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117469
If \((x-i y)^{1 / 3}=a-i b\), then \(\frac{x}{a}+\frac{y}{b}\) is equal to
1 \(-2\left(a^2+b^2\right)\)
2 \(4(a+b)\)
3 \(4(a-b)\)
4 \(4 \mathrm{ab}\)
Explanation:
A Given, \((x-i y)^{1 / 3}=a-i b\) Now cubing both side \(\left\{(x-i y)^{1 / 3}\right\}^3=(a-i b)^3\) \((x-i y)=a^3-(i b)^3-3 a i b(a-i b)\) \(=a^3+i b^3-3 a^2 i b+\left(3 a i^2 b^2\right)\) \(=a^3+i b^3-3 a^2 b i-3 a b^2\) \(=\left(a^3-3 a b^2\right)+i\left(b^3-3 a^2 b\right)\) Comparing the value of \(x\) and \(y\) \(x=a\left(a^2-3 b^2\right) \text { and } y=b\left(b^2-3 a^2\right)\) Then, \(\frac{x}{a}+\frac{y}{b} =\frac{a\left(a^2-3 b^2\right)}{a}+\frac{b\left(b^2-3 a^2\right)}{b}\) \(\frac{x}{a}+\frac{y}{b} =a^2-3 b^2+b^2-3 a^2\) \(=-2 a^2-2 b^2\) \(\frac{x}{a}+\frac{y}{b} =-2\left(a^2+b^2\right)\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117470
If \((a+i b)(c+i d)(e+i f)(g+i h)=A+i B\), then \(\left(a^2+b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right)=\)
1 \(\mathrm{A}^2+\mathrm{B}^2\)
2 \(\mathrm{A}^2-\mathrm{B}^2\)
3 \(\mathrm{A}^2\)
4 \(\mathrm{B}^2\)
Explanation:
A Given, \((a+i b)(c+i d)(e+i f)(g+i h)=A+i B\) Then conjugate of \(\mathrm{A}+\mathrm{iB}\) is \(\mathrm{A}-\mathrm{iB}\) \(\Rightarrow \quad(\mathrm{a}-\mathrm{ib})(\mathrm{c}-\mathrm{id})(\mathrm{e}-\mathrm{if})(\mathrm{g}-\mathrm{ih})=\mathrm{A}-\mathrm{iB}\) Multiplying (1) and (2), we get \(\left(\mathrm{a}^2+\mathrm{b}^2\right)\left(\mathrm{c}^2+\mathrm{d}^2\right)\left(\mathrm{e}^2+\mathrm{f}^2\right)\left(\mathrm{g}^2+\mathrm{h}^2\right)=\mathrm{A}^2+\mathrm{B}^2\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117471
\(\quad(z+a)(\bar{z}+a)\), where \(a\) is real, is equivalent to
1 \(|z-a|\)
2 \(z^2+a^2\)
3 \(|z+a|^2\)
4 None of these
Explanation:
C \((z+a)(\bar{z}+a)=\text { ? }\) Given that \(a\) is real then, \(\overline{\mathrm{a}}=\mathrm{a}\) \((z+a)(\bar{z}+a)=(z+a)(\overline{z+a})\) We know that, \(z \bar{z}=|z|^2\)So, \(\quad(z+a)(\overline{z+a})=|z+a|^2\)
A Let \(x=(16)^{1 / 4}\) \(x^4=16\) \(\mathrm{x}^4-16=0\) \(\left(\mathrm{x}^2\right)^2-\left(2^2\right)^2=0\) \(\left(x^2-2^2\right)\left(x^2+2^2\right)=0\) \((\mathrm{x}-2)(\mathrm{x}+2)\left(\mathrm{x}^2+4\right)=0\) \(\mathrm{x}= \pm 2\) \(x^2=-4\) \(x= \pm 2 \mathrm{i}\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117469
If \((x-i y)^{1 / 3}=a-i b\), then \(\frac{x}{a}+\frac{y}{b}\) is equal to
1 \(-2\left(a^2+b^2\right)\)
2 \(4(a+b)\)
3 \(4(a-b)\)
4 \(4 \mathrm{ab}\)
Explanation:
A Given, \((x-i y)^{1 / 3}=a-i b\) Now cubing both side \(\left\{(x-i y)^{1 / 3}\right\}^3=(a-i b)^3\) \((x-i y)=a^3-(i b)^3-3 a i b(a-i b)\) \(=a^3+i b^3-3 a^2 i b+\left(3 a i^2 b^2\right)\) \(=a^3+i b^3-3 a^2 b i-3 a b^2\) \(=\left(a^3-3 a b^2\right)+i\left(b^3-3 a^2 b\right)\) Comparing the value of \(x\) and \(y\) \(x=a\left(a^2-3 b^2\right) \text { and } y=b\left(b^2-3 a^2\right)\) Then, \(\frac{x}{a}+\frac{y}{b} =\frac{a\left(a^2-3 b^2\right)}{a}+\frac{b\left(b^2-3 a^2\right)}{b}\) \(\frac{x}{a}+\frac{y}{b} =a^2-3 b^2+b^2-3 a^2\) \(=-2 a^2-2 b^2\) \(\frac{x}{a}+\frac{y}{b} =-2\left(a^2+b^2\right)\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117470
If \((a+i b)(c+i d)(e+i f)(g+i h)=A+i B\), then \(\left(a^2+b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right)=\)
1 \(\mathrm{A}^2+\mathrm{B}^2\)
2 \(\mathrm{A}^2-\mathrm{B}^2\)
3 \(\mathrm{A}^2\)
4 \(\mathrm{B}^2\)
Explanation:
A Given, \((a+i b)(c+i d)(e+i f)(g+i h)=A+i B\) Then conjugate of \(\mathrm{A}+\mathrm{iB}\) is \(\mathrm{A}-\mathrm{iB}\) \(\Rightarrow \quad(\mathrm{a}-\mathrm{ib})(\mathrm{c}-\mathrm{id})(\mathrm{e}-\mathrm{if})(\mathrm{g}-\mathrm{ih})=\mathrm{A}-\mathrm{iB}\) Multiplying (1) and (2), we get \(\left(\mathrm{a}^2+\mathrm{b}^2\right)\left(\mathrm{c}^2+\mathrm{d}^2\right)\left(\mathrm{e}^2+\mathrm{f}^2\right)\left(\mathrm{g}^2+\mathrm{h}^2\right)=\mathrm{A}^2+\mathrm{B}^2\)
MHT CET-2021
Complex Numbers and Quadratic Equation
117471
\(\quad(z+a)(\bar{z}+a)\), where \(a\) is real, is equivalent to
1 \(|z-a|\)
2 \(z^2+a^2\)
3 \(|z+a|^2\)
4 None of these
Explanation:
C \((z+a)(\bar{z}+a)=\text { ? }\) Given that \(a\) is real then, \(\overline{\mathrm{a}}=\mathrm{a}\) \((z+a)(\bar{z}+a)=(z+a)(\overline{z+a})\) We know that, \(z \bar{z}=|z|^2\)So, \(\quad(z+a)(\overline{z+a})=|z+a|^2\)
