A \(a+i b=i^i\) Taking \(\log\) both side - \(\log (\mathrm{a}+\mathrm{ib})=\log \left(\mathrm{i}^{\mathrm{i}}\right)\) \(=\mathrm{i} \log (\mathrm{i})\) \(=\mathrm{i} \log \left(\mathrm{e}^{\mathrm{i} \pi / 2}\right)\left\{\because \mathrm{e}^{\mathrm{i} \pi / 2}=\cos \pi / 2+\mathrm{i} \sin \pi / 2=\mathrm{i}\right\}\) \(=\mathrm{i} \times \mathrm{i} \pi / 2\) \(=i^2 \pi / 2\) \(\log (a+i b)=-\pi / 2\) \(a+i b=e^{-\pi / 2}\) On comparing imaginary part is zero.
Karnataka CET-2007
Complex Numbers and Quadratic Equation
117497
Number of solutions of the equation \(\mathrm{z}^2+ \vert \mathrm{z} \vert ^2=0\) is
1 1
2 2
3 3
4 infinitely many
Explanation:
D \( We have, \mathrm{z}^2+ \vert \mathrm{z} \vert ^2=0, \mathrm{z} \neq 0\) \(\text { Let } \mathrm{z}=\mathrm{x}+\mathrm{iy}\)
\(\Rightarrow \mathrm{x}^2-\mathrm{y}^2+\mathrm{i} 2 \mathrm{xy}+\mathrm{x}^2+\mathrm{y}^2=0\)
\(\Rightarrow 2 \mathrm{x}^2+\mathrm{i} 2 \mathrm{xy}=0 \Rightarrow 2 \mathrm{x}(\mathrm{x}+\mathrm{iy})=0\)
\(\Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}+\mathrm{iy}=0 \quad(\text { not possible) }\)
\(\therefore \mathrm{x}=0 \text { and } \mathrm{z} \neq 0\)
So, y can have any real value.
Hence infinitely many solutions.
B \( We have, \mathrm{i}^{141}+\mathrm{i}^{142}+\mathrm{i}^{143}+\mathrm{i}^{144}\) \(=\mathrm{i}^{140}\left[\mathrm{i}+\mathrm{i}^2+\mathrm{i}^3+\mathrm{i}^4\right]\)
\(=\left(\mathrm{i}^4\right)^{35}[\mathrm{i}-1-\mathrm{i}+1]\)
\(=0\)
\(=0\)
COMEDK-2020
Complex Numbers and Quadratic Equation
117500
The number of solutions of the equation \(\mathrm{x}^2-5 \vert \mathrm{x} \vert +6=0\) is
1 4
2 3
3 2
4 1
Explanation:
A Given equation - \(x^2-5 \vert x \vert +6=0\)
Then,
\(x^2-5 x+6 ; x>0\)
\(x^2+5 x+6 ; x\lt 0\)
From the above equations roots are -
\(x=(2,3) \&(-2,-3)\)Hence, Number of solution of given equation is4.
A \(a+i b=i^i\) Taking \(\log\) both side - \(\log (\mathrm{a}+\mathrm{ib})=\log \left(\mathrm{i}^{\mathrm{i}}\right)\) \(=\mathrm{i} \log (\mathrm{i})\) \(=\mathrm{i} \log \left(\mathrm{e}^{\mathrm{i} \pi / 2}\right)\left\{\because \mathrm{e}^{\mathrm{i} \pi / 2}=\cos \pi / 2+\mathrm{i} \sin \pi / 2=\mathrm{i}\right\}\) \(=\mathrm{i} \times \mathrm{i} \pi / 2\) \(=i^2 \pi / 2\) \(\log (a+i b)=-\pi / 2\) \(a+i b=e^{-\pi / 2}\) On comparing imaginary part is zero.
Karnataka CET-2007
Complex Numbers and Quadratic Equation
117497
Number of solutions of the equation \(\mathrm{z}^2+ \vert \mathrm{z} \vert ^2=0\) is
1 1
2 2
3 3
4 infinitely many
Explanation:
D \( We have, \mathrm{z}^2+ \vert \mathrm{z} \vert ^2=0, \mathrm{z} \neq 0\) \(\text { Let } \mathrm{z}=\mathrm{x}+\mathrm{iy}\)
\(\Rightarrow \mathrm{x}^2-\mathrm{y}^2+\mathrm{i} 2 \mathrm{xy}+\mathrm{x}^2+\mathrm{y}^2=0\)
\(\Rightarrow 2 \mathrm{x}^2+\mathrm{i} 2 \mathrm{xy}=0 \Rightarrow 2 \mathrm{x}(\mathrm{x}+\mathrm{iy})=0\)
\(\Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}+\mathrm{iy}=0 \quad(\text { not possible) }\)
\(\therefore \mathrm{x}=0 \text { and } \mathrm{z} \neq 0\)
So, y can have any real value.
Hence infinitely many solutions.
B \( We have, \mathrm{i}^{141}+\mathrm{i}^{142}+\mathrm{i}^{143}+\mathrm{i}^{144}\) \(=\mathrm{i}^{140}\left[\mathrm{i}+\mathrm{i}^2+\mathrm{i}^3+\mathrm{i}^4\right]\)
\(=\left(\mathrm{i}^4\right)^{35}[\mathrm{i}-1-\mathrm{i}+1]\)
\(=0\)
\(=0\)
COMEDK-2020
Complex Numbers and Quadratic Equation
117500
The number of solutions of the equation \(\mathrm{x}^2-5 \vert \mathrm{x} \vert +6=0\) is
1 4
2 3
3 2
4 1
Explanation:
A Given equation - \(x^2-5 \vert x \vert +6=0\)
Then,
\(x^2-5 x+6 ; x>0\)
\(x^2+5 x+6 ; x\lt 0\)
From the above equations roots are -
\(x=(2,3) \&(-2,-3)\)Hence, Number of solution of given equation is4.
A \(a+i b=i^i\) Taking \(\log\) both side - \(\log (\mathrm{a}+\mathrm{ib})=\log \left(\mathrm{i}^{\mathrm{i}}\right)\) \(=\mathrm{i} \log (\mathrm{i})\) \(=\mathrm{i} \log \left(\mathrm{e}^{\mathrm{i} \pi / 2}\right)\left\{\because \mathrm{e}^{\mathrm{i} \pi / 2}=\cos \pi / 2+\mathrm{i} \sin \pi / 2=\mathrm{i}\right\}\) \(=\mathrm{i} \times \mathrm{i} \pi / 2\) \(=i^2 \pi / 2\) \(\log (a+i b)=-\pi / 2\) \(a+i b=e^{-\pi / 2}\) On comparing imaginary part is zero.
Karnataka CET-2007
Complex Numbers and Quadratic Equation
117497
Number of solutions of the equation \(\mathrm{z}^2+ \vert \mathrm{z} \vert ^2=0\) is
1 1
2 2
3 3
4 infinitely many
Explanation:
D \( We have, \mathrm{z}^2+ \vert \mathrm{z} \vert ^2=0, \mathrm{z} \neq 0\) \(\text { Let } \mathrm{z}=\mathrm{x}+\mathrm{iy}\)
\(\Rightarrow \mathrm{x}^2-\mathrm{y}^2+\mathrm{i} 2 \mathrm{xy}+\mathrm{x}^2+\mathrm{y}^2=0\)
\(\Rightarrow 2 \mathrm{x}^2+\mathrm{i} 2 \mathrm{xy}=0 \Rightarrow 2 \mathrm{x}(\mathrm{x}+\mathrm{iy})=0\)
\(\Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}+\mathrm{iy}=0 \quad(\text { not possible) }\)
\(\therefore \mathrm{x}=0 \text { and } \mathrm{z} \neq 0\)
So, y can have any real value.
Hence infinitely many solutions.
B \( We have, \mathrm{i}^{141}+\mathrm{i}^{142}+\mathrm{i}^{143}+\mathrm{i}^{144}\) \(=\mathrm{i}^{140}\left[\mathrm{i}+\mathrm{i}^2+\mathrm{i}^3+\mathrm{i}^4\right]\)
\(=\left(\mathrm{i}^4\right)^{35}[\mathrm{i}-1-\mathrm{i}+1]\)
\(=0\)
\(=0\)
COMEDK-2020
Complex Numbers and Quadratic Equation
117500
The number of solutions of the equation \(\mathrm{x}^2-5 \vert \mathrm{x} \vert +6=0\) is
1 4
2 3
3 2
4 1
Explanation:
A Given equation - \(x^2-5 \vert x \vert +6=0\)
Then,
\(x^2-5 x+6 ; x>0\)
\(x^2+5 x+6 ; x\lt 0\)
From the above equations roots are -
\(x=(2,3) \&(-2,-3)\)Hence, Number of solution of given equation is4.
A \(a+i b=i^i\) Taking \(\log\) both side - \(\log (\mathrm{a}+\mathrm{ib})=\log \left(\mathrm{i}^{\mathrm{i}}\right)\) \(=\mathrm{i} \log (\mathrm{i})\) \(=\mathrm{i} \log \left(\mathrm{e}^{\mathrm{i} \pi / 2}\right)\left\{\because \mathrm{e}^{\mathrm{i} \pi / 2}=\cos \pi / 2+\mathrm{i} \sin \pi / 2=\mathrm{i}\right\}\) \(=\mathrm{i} \times \mathrm{i} \pi / 2\) \(=i^2 \pi / 2\) \(\log (a+i b)=-\pi / 2\) \(a+i b=e^{-\pi / 2}\) On comparing imaginary part is zero.
Karnataka CET-2007
Complex Numbers and Quadratic Equation
117497
Number of solutions of the equation \(\mathrm{z}^2+ \vert \mathrm{z} \vert ^2=0\) is
1 1
2 2
3 3
4 infinitely many
Explanation:
D \( We have, \mathrm{z}^2+ \vert \mathrm{z} \vert ^2=0, \mathrm{z} \neq 0\) \(\text { Let } \mathrm{z}=\mathrm{x}+\mathrm{iy}\)
\(\Rightarrow \mathrm{x}^2-\mathrm{y}^2+\mathrm{i} 2 \mathrm{xy}+\mathrm{x}^2+\mathrm{y}^2=0\)
\(\Rightarrow 2 \mathrm{x}^2+\mathrm{i} 2 \mathrm{xy}=0 \Rightarrow 2 \mathrm{x}(\mathrm{x}+\mathrm{iy})=0\)
\(\Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}+\mathrm{iy}=0 \quad(\text { not possible) }\)
\(\therefore \mathrm{x}=0 \text { and } \mathrm{z} \neq 0\)
So, y can have any real value.
Hence infinitely many solutions.
B \( We have, \mathrm{i}^{141}+\mathrm{i}^{142}+\mathrm{i}^{143}+\mathrm{i}^{144}\) \(=\mathrm{i}^{140}\left[\mathrm{i}+\mathrm{i}^2+\mathrm{i}^3+\mathrm{i}^4\right]\)
\(=\left(\mathrm{i}^4\right)^{35}[\mathrm{i}-1-\mathrm{i}+1]\)
\(=0\)
\(=0\)
COMEDK-2020
Complex Numbers and Quadratic Equation
117500
The number of solutions of the equation \(\mathrm{x}^2-5 \vert \mathrm{x} \vert +6=0\) is
1 4
2 3
3 2
4 1
Explanation:
A Given equation - \(x^2-5 \vert x \vert +6=0\)
Then,
\(x^2-5 x+6 ; x>0\)
\(x^2+5 x+6 ; x\lt 0\)
From the above equations roots are -
\(x=(2,3) \&(-2,-3)\)Hence, Number of solution of given equation is4.
