Question

If z is any complex number where |z|=1,show that using the Argand plane 1<= |z+2|<=3 and (-pi/6)<= arg(z+2)<= (pi/6)...

I know if we take z=x+iy
|z|= root(x^2+y^2)

so |z+2|=root[(x+2)^2 +y^2]

I don't see a way to proceed

Answer 1

we know that
|z+2| <= |z|+|2| = 1+2 = 3

use similar logic on the other side.

Answer 2

Thank you Steve!