Question

Find the least value of |z-1| if |z+3-i|=2

Answer 1

basically it is circle
|z+3-i|=|z-(-3+i)|=2
with radius= 2 and centre at (-3,i)

Answer 2

|z-1| implies circle with centre at (1,0)

Answer 3

let z= x+iy
|x+iy+3-i|=|x+3+i(y-1)|=2
now taking its magnitude
\[\sqrt{(x+3)^{2}+(y-1)^{2}}=2\]

Answer 4

or
\[(x+3){^2}+(y-1)^{2}=4\]

Answer 5

|z-1|=|x+iy-1|
=\[\sqrt{(x-1){2}+y^2}\]

Answer 6

now convert first equation in the below given

Answer 7

What equation am I converting?

Answer 8

(x+3)^2+(y-1)^2=4 to (x-1)^2+y^2

Answer 9

I don't understand :S

Answer 10

we had the value of (x+3)^2+(y-1)^2
and need to find min of (x-1)^2+y^2