Find the range of real number α for which the equation\[z + \alpha |z - 1| + 2i = 0\], \[z = x + iy\] has a solution. Find the solution.
A. $x = \dfrac{5}{2},\,y = - 2$
B. $x = - 2,\,y = \dfrac{5}{2}$
C. $x = - \dfrac{5}{2},y = 2$
D.$x = 2,\,y = - \dfrac{5}{2}$
Answer
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Hint: Use \[z = x + iy,\,\,|z| = \sqrt {{x^2} + {y^2}} \]to find solution of the equation.
Complete step by step answer:
(1) Given equation is
\[z + \alpha |z - 1| + 2i = 0\]
Here, \[z = x + iy\]
(2) \[x + iy + \alpha |(x + iy) - 1| + 2i = 0\]
\[x + iy + \alpha |(x - 1) + (iy)| + 2i = 0\]
(3) Using mode property
\[x + iy + \alpha \sqrt {{{(x - 1)}^2} + {{(y)}^2}} + 2i = 0\]
Now we will shift the value of\[\alpha \sqrt {{{(x - 1)}^2} + {{(y)}^2}} \]to the right side,
(4) \[x + (y + 2)i = - \alpha \sqrt {{{(x - 1)}^2} + {y^2}} \]
Squaring both sides, we will get
\[{\{ x + (y + 2)i\} ^2} = {\alpha ^2}{\left[ {\sqrt {{{(x - 1)}^2} + {y^2}} } \right]^2}\]
\[{x^2} + {(y + 2)^2} + 2x(y + 2)i = {\alpha ^2}\left[ {{{(x - 1)}^2} + {y^2}} \right]\]
\[ \Rightarrow \] Equating real and imaginary part from both sides
Imaginary part:
\[
2x(y + 2) = 0 \\
\Rightarrow x = 0,\,\,y = - 2 \\
\]
Real part, \[{x^2} + {(y + 2)^2} = {\alpha ^2}|{(x - 1)^2} + {y^2}|\]
Taking \[y = - 2\] in the real part for complex number, we get
\[{x^2} + {( - 2 + 2)^2} = {\alpha ^2}\left[ {{{(x - 1)}^2} + 4} \right]\]
Using algebraic identity: \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\], we will get
$ \Rightarrow {x^2} + 0 = {\alpha ^2}({x^2} + 1 - 2x + 4)$
\[ \Rightarrow ({x^2} - 2x + 1 + 4) = \dfrac{{{x^2}}}{{{\alpha ^2}}}\]
\[ \Rightarrow {x^2} - 2x + 5 = \dfrac{{{x^2}}}{{{\alpha ^2}}}\]
\[\]\[ \Rightarrow {x^2} - 2x + 5 - \dfrac{{{x^2}}}{{{\alpha ^2}}} = 0\]
\[ \Rightarrow {x^2}\left( {1 - \dfrac{1}{{{\alpha ^2}}}} \right) - 2x + 5 = 0\]
For real \[x,D \geqslant 0\]
\[{b^2} - 4ac \geqslant 0\]
\[ \Rightarrow 4 - 4\left( {1 - \dfrac{1}{{{\alpha ^2}}}} \right)5 \geqslant 0\]
\[ \Rightarrow 4 - 20\left( {1 - \dfrac{1}{{{\alpha ^2}}}} \right) \geqslant 0\]
\[ \Rightarrow 4 - 20 + \dfrac{{20}}{{{\alpha ^2}}} \geqslant 0\]
\[ \Rightarrow - 16 + \dfrac{{20}}{{{\alpha ^2}}} \geqslant 0\]
\[ \Rightarrow \dfrac{{20}}{{{\alpha ^2}}} \geqslant 16\]
\[ \Rightarrow 16{\alpha ^2} < 20\]
\[ \Rightarrow {\alpha ^2} < \dfrac{{20}}{{16}} = \dfrac{5}{4}\]
\[ \Rightarrow {\alpha ^2}\alpha \dfrac{5}{4}\]
\[\therefore \alpha \in \left( {\sqrt {\dfrac{{ - 5}}{4}} ,\sqrt {\dfrac{5}{4}} } \right)\]
Or\[\left( { - \dfrac{{\sqrt 5 }}{2},\dfrac{{ + \sqrt 5 }}{2}} \right)\]
(5) Hence, range of real α is \[\left( {\dfrac{{ - \sqrt 5 }}{2},\dfrac{{\sqrt 5 }}{2}} \right)\]
(6) Therefore, the required solution of the \[x = \dfrac{5}{2},\,\,y = - 2\]
Note: The range of a function is the set of outputs. The function achieves when it is applied to its whole set of outputs. A function relates an input to an output. The range is the set of objects that actually come out of the machine when you feed it with all the inputs.
Complete step by step answer:
(1) Given equation is
\[z + \alpha |z - 1| + 2i = 0\]
Here, \[z = x + iy\]
(2) \[x + iy + \alpha |(x + iy) - 1| + 2i = 0\]
\[x + iy + \alpha |(x - 1) + (iy)| + 2i = 0\]
(3) Using mode property
\[x + iy + \alpha \sqrt {{{(x - 1)}^2} + {{(y)}^2}} + 2i = 0\]
Now we will shift the value of\[\alpha \sqrt {{{(x - 1)}^2} + {{(y)}^2}} \]to the right side,
(4) \[x + (y + 2)i = - \alpha \sqrt {{{(x - 1)}^2} + {y^2}} \]
Squaring both sides, we will get
\[{\{ x + (y + 2)i\} ^2} = {\alpha ^2}{\left[ {\sqrt {{{(x - 1)}^2} + {y^2}} } \right]^2}\]
\[{x^2} + {(y + 2)^2} + 2x(y + 2)i = {\alpha ^2}\left[ {{{(x - 1)}^2} + {y^2}} \right]\]
\[ \Rightarrow \] Equating real and imaginary part from both sides
Imaginary part:
\[
2x(y + 2) = 0 \\
\Rightarrow x = 0,\,\,y = - 2 \\
\]
Real part, \[{x^2} + {(y + 2)^2} = {\alpha ^2}|{(x - 1)^2} + {y^2}|\]
Taking \[y = - 2\] in the real part for complex number, we get
\[{x^2} + {( - 2 + 2)^2} = {\alpha ^2}\left[ {{{(x - 1)}^2} + 4} \right]\]
Using algebraic identity: \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\], we will get
$ \Rightarrow {x^2} + 0 = {\alpha ^2}({x^2} + 1 - 2x + 4)$
\[ \Rightarrow ({x^2} - 2x + 1 + 4) = \dfrac{{{x^2}}}{{{\alpha ^2}}}\]
\[ \Rightarrow {x^2} - 2x + 5 = \dfrac{{{x^2}}}{{{\alpha ^2}}}\]
\[\]\[ \Rightarrow {x^2} - 2x + 5 - \dfrac{{{x^2}}}{{{\alpha ^2}}} = 0\]
\[ \Rightarrow {x^2}\left( {1 - \dfrac{1}{{{\alpha ^2}}}} \right) - 2x + 5 = 0\]
For real \[x,D \geqslant 0\]
\[{b^2} - 4ac \geqslant 0\]
\[ \Rightarrow 4 - 4\left( {1 - \dfrac{1}{{{\alpha ^2}}}} \right)5 \geqslant 0\]
\[ \Rightarrow 4 - 20\left( {1 - \dfrac{1}{{{\alpha ^2}}}} \right) \geqslant 0\]
\[ \Rightarrow 4 - 20 + \dfrac{{20}}{{{\alpha ^2}}} \geqslant 0\]
\[ \Rightarrow - 16 + \dfrac{{20}}{{{\alpha ^2}}} \geqslant 0\]
\[ \Rightarrow \dfrac{{20}}{{{\alpha ^2}}} \geqslant 16\]
\[ \Rightarrow 16{\alpha ^2} < 20\]
\[ \Rightarrow {\alpha ^2} < \dfrac{{20}}{{16}} = \dfrac{5}{4}\]
\[ \Rightarrow {\alpha ^2}\alpha \dfrac{5}{4}\]
\[\therefore \alpha \in \left( {\sqrt {\dfrac{{ - 5}}{4}} ,\sqrt {\dfrac{5}{4}} } \right)\]
Or\[\left( { - \dfrac{{\sqrt 5 }}{2},\dfrac{{ + \sqrt 5 }}{2}} \right)\]
(5) Hence, range of real α is \[\left( {\dfrac{{ - \sqrt 5 }}{2},\dfrac{{\sqrt 5 }}{2}} \right)\]
(6) Therefore, the required solution of the \[x = \dfrac{5}{2},\,\,y = - 2\]
Note: The range of a function is the set of outputs. The function achieves when it is applied to its whole set of outputs. A function relates an input to an output. The range is the set of objects that actually come out of the machine when you feed it with all the inputs.
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