Question

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 1

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.