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
we know that
|z+2| <= |z|+|2| = 1+2 = 3
use similar logic on the other side.
Thank you Steve!
To solve this problem, you can approach it by breaking it down into two parts: the modulus inequality and the argument inequality.
For the modulus inequality, you need to show that 1 ≤ |z+2| ≤ 3. Let's start with the lower bound.
1 ≤ |z+2|
Square both sides:
1 ≤ (x+2)^2 + y^2
We know that |z| = 1, so substituting z = x+iy, we have:
1 ≤ (x^2+y^2) + 4x + 4 + y^2
1 - (x^2+y^2) ≤ 4x + y^2 + 4
- (x^2+y^2) ≤ 4x + y^2 + 3
Since the left-hand side is negative and the right-hand side is positive, this inequality is always satisfied.
Now let's move on to the upper bound.
|z+2| ≤ 3
(x+2)^2 + y^2 ≤ 9
Rewrite this in terms of x and y:
x^2 + 4x + 4 + y^2 ≤ 9
x^2 + y^2 + 4x - 5 ≤ 0
This is a quadratic equation in x and y. We can rewrite it as:
(x+2)^2 + y^2 - 1 ≤ 0
Now, we have a circle centered at (-2, 0) with radius 1. Any point inside or on this circle satisfies the inequality.
For the argument inequality, you need to show that (-π/6) ≤ arg(z+2) ≤ (π/6).
Let's start with the lower bound:
(-π/6) ≤ arg(z+2)
Take the tangent of both sides:
tan(-π/6) ≤ tan(arg(z+2))
The tangent function is positive in the interval (-π/2, π/2), so we have:
- tan(π/6) ≤ tan(arg(z+2))
-1/√3 ≤ tan(arg(z+2))
This means that the argument of (z+2) must lie in the interval (-π/6, π/6].
For the upper bound, follow the same steps:
tan(π/6) ≥ tan(arg(z+2))
1/√3 ≥ tan(arg(z+2))
Therefore, the argument of (z+2) must lie in the interval [-π/6, π/6].
By considering both the modulus inequality and the argument inequality, you can conclude that 1 ≤ |z+2| ≤ 3 and (-π/6) ≤ arg(z+2) ≤ (π/6) hold for any complex number z where |z| = 1 in the Argand plane.