Complex Numbers and Quadratic Equations Test 23

$$f\left( z \right) =\dfrac { 1-{ z }^{ 3 } }{ 1-z } =\dfrac { \left( 1-z \right) \left( 1+z+{ z }^{ 2 } \right)  }{ \left( 1-z \right)  } $$
         $$=1+z+{ z }^{ 2 }$$
Put $$z=x+iy$$, we get
$$f\left( z \right) =1+x+iy+{ \left( x+iy \right)  }^{ 2 }$$
    $$=1+x+iy+{ x }^{ 2 }+2xyi+{ i }^{ 2 }{ y }^{ 2 }$$
    $$=\left( 1+x+{ x }^{ 2 }-{ y }^{ 2 } \right) +i\left( y+2xy \right) $$
$$\Rightarrow f\left( \overline { z }  \right) =\left( 1+x+{ x }^{ 2 }-{ y }^{ 2 } \right) -i\left( y+2xy \right) $$
Now, $$Re\left\{ \overline { f\left( z \right)  }  \right\} =0$$
$$\Rightarrow 1+x+{ x }^{ 2 }-{ y }^{ 2 }=0$$
$$\Rightarrow { x }^{ 2 }-{ y }^{ 2 }+x+1=0$$

$$Re\left( z \right) =\left| z-2 \right| $$
$$Re\left( w \right) =\left| w-z \right| $$
$$arg\left( z-w \right) =\cfrac { \pi }{ 3 } $$

Let $$w={ z }^{ 2 }{ e }^{ z-i }$$ 
Put $$z=1+i$$
$$\Rightarrow w={ \left( 1+i \right)  }^{ 2 }{ e }^{ 1+i-i }=\left( 1+2i+{ i }^{ 2 } \right) e$$
$$=\left( 1+2i-1 \right) e=2ei$$
Now, argument of $$w=arg\left( w \right) $$
$$=\tan ^{ -1 }{ \left( \dfrac { 2e }{ 0 }  \right)  } =\tan ^{ -1 }{ \infty  } =\dfrac { \pi  }{ 2 } $$

Given equation is $$2px^{2} + (2p + q) x + q = 0$$ ........ $$(i)$$

Given : $$iz^{3} + z^{2} - z + i = 0$$
Dividing both sides by $$i$$ and using $$\dfrac {1}{i} = -i$$.
We have,
$$z^{3} - iz^{2} + iz + 1 = 0$$
$$\Rightarrow z^{2}(z - i) + i(z - i) = 0$$ ...... $$(\because i^{2} = -1)$$
$$\Rightarrow (z - i)(z^{2} + i) = 0$$
$$\therefore z = i$$ or $$z^{2} = -i$$
$$\therefore |z| = |i| = 1$$
and $$|z|^{2}=|z^{2}| = |-i| = 1$$
$$\therefore |z| =1$$.

$$arg(z)=arg(Numerator)-arg(Denominator)$$
$$=\tan ^{ -1 }{ \left| \cfrac { \cos { \cfrac { \pi  }{ 3 }  }  }{ 1+\sin { \cfrac { \pi  }{ 3 }  }  }  \right|  } +\tan ^{ -1 }{ \left| \cfrac { \cos { \cfrac { \pi  }{ 3 }  }  }{ 1+\sin { \cfrac { \pi  }{ 3 }  }  }  \right|  } $$
$$=2\tan ^{ -1 }{ \left| \cfrac { \cos { \cfrac { \pi  }{ 3 }  }  }{ 1+\sin { \cfrac { \pi  }{ 3 }  }  }  \right|  } $$
$$=2\tan ^{ -1 }{ (2-\sqrt { 3 }  } )=2\times {15}^{o}=\cfrac { \pi  }{ 6 } $$

The value of $$ \sum _{ k=0 }^{ n }{ (i^k + i^{k+1} ) } =  \sum _{ k=0 }^{ n }{ i^k ( {i +1} ) }  $$ is, 
$$ = (i+1) \left[ \dfrac {1(1-i^{n+1})} {(1-i)} \right] \times \dfrac {1+i}{1+i} $$
$$ = \dfrac {(1+i)^2 }{1+1} = \dfrac {(1-1+2i)(1-i^{n+1} )}{2} $$
$$ = \dfrac {2i}{2} ( 1 - i^{n+1} ) $$

Given, $$z_{1} + i \,\overline {(z_{2})} = 0$$
$$\Rightarrow z_{1} = (-i) \overline {z_{2}}$$
Taking argument on both sides, we get
$$arg (z_{1}) = arg \left \{(-i) \cdot \overline {z_{2}}\right \}$$
$$\Rightarrow  arg (z_{1}) = arg (-i) + arg (\overline {z_{2}})$$ (by property)
$$\Rightarrow arg(z_{1}) - arg (\overline {z_{2}}) = \tan^{-1} \left (\dfrac {-1}{0}\right ) = -\dfrac {\pi}{2}$$
$$\Rightarrow arg (z_{1}) + arg (z_{2}) = \dfrac {-\pi}{2} ..... (i)$$
$$[\because arg (\overline {z}) = -arg (z)]$$
and $$arg (\overline {z_{1}}z_{2}) = \dfrac {\pi}{3}$$
$$\Rightarrow arg (\overline {z_{1}}) + arg (z_{2}) = \dfrac {\pi}{3}$$
$$\Rightarrow -arg (z_{1}) + arg (z_{2}) = \dfrac {\pi}{3}... (ii)$$
On adding Eqs. (i) and (ii), we get
$$2 arg (z_{2}) = \dfrac {-\pi}{6}$$
$$\Rightarrow arg (z_{2}) = \dfrac {-\pi}{12}$$
Therefore, from Eq. (i),
$$arg (z_{1}) = \dfrac {-\pi}{2} + \dfrac {\pi}{12} = \dfrac {-5\pi}{12}$$
$$\Rightarrow -arg (z_{1}) = \dfrac {5\pi}{12}$$
$$\Rightarrow arg (z_{1}) = \dfrac {5\pi}{12}$$.

We know that, $$\left| z-{ z }_{ 1 } \right| >\left| z-{ z }_{ 2 } \right| $$ represents the region on right side of perpendicular bisector of $${z}^{1}$$ and $${z}^{2}$$.
Thus, the given inequality $$\left| z-2 \right| >\left| z-4 \right| $$ represents the region on the right side of perpendicular bisector of $$2$$ and $$4$$.
$$\Rightarrow Re(z)>3$$ and $$\text{Im}(z)\in R$$