Let's tackle this problem step-by-step.
### Part (i) Finding [tex]\(\frac{u}{v}\)[/tex] in the form [tex]\(x + iy\)[/tex]:
Given:
[tex]\[ u = p + 2i \][/tex]
[tex]\[ v = 1 - 2i \][/tex]
We need to find [tex]\(\frac{u}{v}\)[/tex] and express it in the form [tex]\(x + iy\)[/tex].
First, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of [tex]\(1 - 2i\)[/tex] is [tex]\(1 + 2i\)[/tex]:
[tex]\[ \frac{u}{v} = \frac{(p + 2i)}{(1 - 2i)} \cdot \frac{(1 + 2i)}{(1 + 2i)} \][/tex]
This simplifies to:
[tex]\[ \frac{(p + 2i)(1 + 2i)}{(1 - 2i)(1 + 2i)} \][/tex]
Now, we evaluate the denominator:
[tex]\[ (1 - 2i)(1 + 2i) = 1^2 - (2i)^2 = 1 - 4(-1) = 1 + 4 = 5 \][/tex]
Next, we evaluate the numerator using the distributive property:
[tex]\[ (p + 2i)(1 + 2i) = p \cdot 1 + p \cdot 2i + 2i \cdot 1 + 2i \cdot 2i \][/tex]
[tex]\[ = p + 2pi + 2i + 4i^2 \][/tex]
Since [tex]\(i^2 = -1\)[/tex], we have:
[tex]\[ = p + 2pi + 2i + 4(-1) \][/tex]
[tex]\[ = p + 2pi + 2i - 4 \][/tex]
[tex]\[ = (p - 4) + 2pi + 2i \][/tex]
Since both [tex]\(p\)[/tex] and [tex]\(2i\)[/tex] are real terms, they can combine:
[tex]\[ = (p - 4) + 2(p + 1)i \][/tex]
Therefore:
[tex]\[ \frac{u}{v} = \frac{(p - 4) + 2(p + 1)i}{5} \][/tex]
[tex]\[ = \frac{p - 4}{5} + \frac{2(p + 1)i}{5} \][/tex]
Thus, in the form [tex]\(x + iy\)[/tex]:
[tex]\[ x = \frac{p - 4}{5} \][/tex]
[tex]\[ y = \frac{2(p + 1)}{5} \][/tex]
### Part (ii) Given [tex]\(\left|\frac{1}{u}\right| = 13\)[/tex], find the possible values of [tex]\(p\)[/tex]:
We know that:
[tex]\[ \left|\frac{1}{u}\right| = 13 \][/tex]
We also know the magnitude of a complex number [tex]\(u = p + 2i\)[/tex] is given by:
[tex]\[ |u| = \sqrt{p^2 + 2^2} = \sqrt{p^2 + 4} \][/tex]
The magnitude of [tex]\(\frac{1}{u}\)[/tex] is:
[tex]\[ \left|\frac{1}{u}\right| = \frac{1}{|u|} = 13 \][/tex]
This implies:
[tex]\[ \frac{1}{|u|} = 13 \][/tex]
[tex]\[ |u| = \frac{1}{13} \][/tex]
Thus:
[tex]\[ \sqrt{p^2 + 4} = \frac{1}{13} \][/tex]
Squaring both sides, we get:
[tex]\[ p^2 + 4 = \left(\frac{1}{13}\right)^2 \][/tex]
[tex]\[ p^2 + 4 = \frac{1}{169} \][/tex]
Then solving for [tex]\(p^2\)[/tex]:
[tex]\[ p^2 + 4 = \frac{1}{169} \][/tex]
[tex]\[ p^2 = \frac{1}{169} - 4 \][/tex]
[tex]\[ p^2 = \frac{1}{169} - \frac{676}{169} \][/tex]
[tex]\[ p^2 = \frac{1 - 676}{169} \][/tex]
[tex]\[ p^2 = \frac{-675}{169} \][/tex]
Since the square of a number cannot be negative, there are no possible real integer solutions for [tex]\(p\)[/tex].
Hence, based on the given conditions, there are no possible integer values of [tex]\(p\)[/tex].
