Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
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].
### 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].
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
