91Ó°ÊÓ

To calculate the modulus of the given complex number z, we can rewrite the number in the form $$a+bi$$ and use the formula $$|z|= \sqrt{a^2+b^2}$$. First, let's rewrite z by multiplying the numerator and denominator by the conjugate of the denominator.

Multiply the numerator and denominator by the conjugate of $$2-i$$, which is $$2+i$$. $$ z= \frac{2+i}{2-i} \cdot \frac{2+i}{2+i} = \frac{(2+i)(2+i)}{(2-i)(2+i)} $$

Simplify the numerator and denominator: $$ z = \frac{4+2i+2i+i^2}{4+ i^2} = \frac{4+ 4i - 1}{4 - 1} = 1 + 2i $$

Now that we have z in the form $$1 + 2i$$, we can compute its modulus using the formula $$|z| = \sqrt{a^2+b^2}$$: $$ |z| = \sqrt{(1)^2 + (2)^2} = \sqrt{1+4} = \sqrt{5} $$ This is not equal to 1, so we made a calculation error in previous steps. Going back, we realize that we should have simplified Step 3 using $$i^2=-1$$: $$ z = \frac{4+ 4i - 1}{4 + 1} = \frac{3+ 4i}{5} = \frac{3}{5} + \frac{4i}{5} $$ Now, recalculate the modulus of z: $$ |z| = \sqrt{(\frac{3}{5})^2 + (\frac{4}{5})^2} = \sqrt{\frac{9}{25}+\frac{16}{25}} = \sqrt{\frac{25}{25}} = 1 $$ The modulus of z is indeed equal to 1.

If z is an n-th root of unity, it will satisfy the equation $$z^n = 1$$ for some positive integer n. We can write z in polar form and use De Moivre's Theorem to check the equality.

To find the polar form, we need the magnitude |z| and the argument (angle) $$\theta$$. We already know |z| = 1. To find the angle, we use the formula $$\tan^{-1}(\frac{b}{a})$$ where a and b are the real and imaginary parts of z, respectively. $$ \theta = \tan^{-1}(\frac{4}{3}) $$ So, the polar form of z is $$z=1(\cos{\theta}+i\sin{\theta})$$.

Using De Moivre's Theorem, we raise z to the power of n and check if it is equal to 1. $$ z^n = (\cos{\theta}+i\sin{\theta})^n = \cos{n\theta}+i\sin{n\theta} $$ For z^n to be equal to 1, the angle $$n\theta$$ must be a multiple of $$2\pi$$, i.e., $$n\theta = 2k\pi$$ for some integer k. However, as $$\theta = \tan^{-1}(\frac{4}{3}) $$, it is clear that $$\theta$$ is not a rational multiple of $$\pi$$ (since $$\tan^{-1}(x)$$ is a transcendental function and the value $$\frac{4}{3}$$ is rational, meaning that $$\tan^{-1}(\frac{4}{3})$$ is not an integer multiple of $$\pi$$). Therefore, it is not possible to find an integer n such that $$n\theta = 2k\pi$$. This implies that z is not an n-th root of unity for any positive integer n. In conclusion, we have proven that the complex number $$z = \frac{2+i}{2-i}$$ has a modulus equal to 1, and it is not an n-th root of unity for any positive integer n.