 2022-01-17

If $|z-6|<|z-2|$, what is its solution given given by? nick1337

Expert

Step 1
The inequality is the same as
$|z-6{|}^{2}<|z-2{|}^{2}$
hence as
$\left(z-6\right)\left(\overline{z}-6\right)<\left(z-2\right)\left(\overline{z}-2\right)$
Expanding and simplifying we get
$4\left(z+\overline{z}\right)>32$
that is
$\frac{z+\overline{z}}{2}>4$
So the solutions are all complex numbers whose real part is greater than 4 star233

Expert

Step 1
Given
$|z-6|<|z-2|⇒\phantom{\rule{0ex}{0ex}}\sqrt{\left(x-6{\right)}^{2}+{y}^{2}}<\sqrt{\left(x-2{\right)}^{2}+{y}^{2}}$
or
$\left(x-6{\right)}^{2}+{y}^{2}<\left(x-2{\right)}^{2}+y⇒8x>32$or
$x>4$ alenahelenash

Expert

Step 1 $z>4$ detalis There are 3 cases which are: First when $z-6\ge 0$ then $z-2>0$ and $|z-6|<|z-2|$ becomes $z-6 subtract z gives $-6<-2$ So $|z-6|<|z-2|$ is true when $z\ge 6$ Second For $6>z\ge 2$ $|z-6|<|z-2|$ becomes $6-z $8<2z$ $4 Combining the first two cases $z>4$ and $z\ge 6$ $z>4$ And finally for case three $z<2$ $|z-6|<|z-2|$ becomes $6-z<2-z$ add z $6<2$ which is false