Complex getting more complex!

${ z }^{ 5 }=\cfrac { 1\pm \sqrt { 1+4\times 992 } }{ 2 } =32,-31$

Now we will use only common sense,

$Let\quad { z }^{ 5 }=32\quad have\quad m\quad root\quad for\quad which\quad Re\left( z \right) <0\\ \\ \Rightarrow no.\quad of\quad roots\quad of\quad { z }^{ 5 }=-31\quad for\quad which\quad Re\left( z \right) >0\quad \\ \\ =m\quad \left( why? \right) \\ \\ \Rightarrow no.\quad of\quad roots\quad of\quad { z }^{ 5 }=-31\quad for\quad which\quad Re\left( z \right) <0\quad \\ \\ =\left( 5-m \right) \\ \\ \therefore Total\quad number\quad roots\quad for\quad which\quad Re\left( z \right) <0\\ \\ =m+(5-m)=5\\$

Note-This method worked only because there is no possible root for which Re(z)=0

Re(z) is real part of z . ( x of x+iy )