There exist two complex numbers $c$, say $c_1$ and $c_2$, so that $3 + 2i$, $6 + i$, and $c$ form the vertices of an equilateral triangle. Find the product $c_1 c_2$ in rectangular form.

Answers

Answer 1

The product $c_1 c_2$ in rectangular frame is [tex]\$10 e^{i(2\theta \pm \frac{pi}{3})}\$.[/tex]

How to find the product $c_1 c_2$ in rectangular form

Let $c_1$ and $c_2$ be the two complex numbers that frame an even triangle with $3+2i$ and $6+i$.

The dispose of between two complex numbers $a$ and $b$ is given by $|b-a|$. Since the triangle is even, the wipe out someplace in the scope of $3+2i$ and $6+i$ ought to be equivalent to the underlying venture with to the kill among $3+2i$ and $c$, conjointly make back the underlying speculation with to the take out among $6+i$ and $c$.

So, we have the following conditions:

$|6 + i - (3 + 2i)| = |c - (3 + 2i)|$

$|6 + i - (3 + 2i)| = |c - (6 + i)|$

Tackling these conditions, we get:

$|3 - i| = |c - (3 + 2i)|$

$|3 - i| = |c - 6 - i|$

Presently, let's discover the esteem of $c$:

$|c - (3 + 2i)| = |3 - i|$

[tex]\$|c - (3 + 2i)| = \sqrt{(3-0)^{2} + (-1-0)^{2} = \sqrt{10}\$[/tex]

So, [tex]\$c - (3 + 2i) = \pm \sqrt{10} e^{i\theta}\$[/tex], where [tex]\$\theta\$[/tex] is the point between the positive genuine hub and the line interfacing $3+2i$ and $c$?

Presently, let's discover the esteem of $c$ utilizing the moment condition:

$|c - (6 + i)| = |3 - i|$

[tex]\$|c - (6 + i)| = \sqrt{(3-0)^{2} + (-1-0)^{2} = \sqrt{10}\$[/tex]

So, [tex]\$c - (6 + i) = \pm \sqrt{10} e^{i\theta'}\$[/tex], where [tex]\$\theta'\$[/tex] is the point between the positive genuine hub and the line interfacing $6+i$ and $c$?

Since the triangle is equilateral, $theta$  [tex]\$\theta'\$[/tex] must vary by [tex]\$\pm[/tex] [tex]\frac{\pi}{3}\$.[/tex]

Presently, let's discover the item $c_1 c_2$:

$c_1 c_2 = (c - (3 + 2i))(c - (6 + i))$

$c_1 c_2 =[tex](\pm \sqrt{10} e^{i\theta})(\pm \sqrt{10} e^{i(\theta \pm \frac{\pi}{3})})\$[/tex]

$c_1 c_2 = [tex]10 e^{i\theta} e^{i(\theta \pm \frac{\pi}{3})}\$[/tex]

$c_1 c_2 = [tex]10 e^{i(2\theta \pm \frac{\pi}{3})}\$[/tex]

So, the item $c_1 c_2$ in rectangular frame is [tex]\$10 e^{i(2\theta \pm \frac{\pi}{3})}\$.[/tex]

Learn more about rectangular form here:

https://brainly.com/question/31002441

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Part A
This year, water usage has increased by 20% for both regular days and special events.
LAST YEAR'S WATER USAGE:
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Per special event - 18.5 gallons
Total for the year - 13,616 gallons
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Answers

Answer:

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Step-by-step explanation:

The given parameters are;

The increase in water usage for both regular days and special event for the current year = 20%

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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