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

Answered question

2022-01-17

If |z6|<|z2|, what is its solution given given by?

Answer & Explanation

nick1337

nick1337

Expert2022-01-17Added 777 answers

Step 1
The inequality is the same as
|z6|2<|z2|2
hence as
(z6)(z¯6)<(z2)(z¯2)
Expanding and simplifying we get
4(z+z¯)>32
that is
z+z¯2>4
So the solutions are all complex numbers whose real part is greater than 4

star233

star233

Skilled2022-01-17Added 403 answers

Step 1
Given
|z6|<|z2|(x6)2+y2<(x2)2+y2
or
(x6)2+y2<(x2)2+y8x>32or
x>4

alenahelenash

alenahelenash

Expert2022-01-24Added 556 answers

Step 1 z>4 detalis There are 3 cases which are: First when z60 then z2>0 and |z6|<|z2| becomes z6<z2 subtract z gives 6<2 So |z6|<|z2| is true when z6 Second For 6>z2 |z6|<|z2| becomes 6z<z2 8<2z 4<z Combining the first two cases z>4 and z6 z>4 And finally for case three z<2 |z6|<|z2| becomes 6z<2z add z 6<2 which is false

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