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3 years ago

If we solve an equation for a variable x what should our final equation look like ?

Mathematics

1 answer:

Nesterboy [21]3 years ago

6 0

You can solve an easy equation in your head by using the multiplication table.

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Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

Read 2 more answers

Akron Cinema sells an average of 500 tickets on Mondays, with a standard deviation of 50 tickets. If a simple random sample is t

Paha777 [63]

Answer:

Step-by-step explanation:

Assuming the number of tickets sales from Mondays is normally distributed. the formula for normal distribution would be applied. It is expressed as

z = (x - u)/s

Where

x = ticket sales from monday

u = mean amount of ticket

s = standard deviation

From the information given,

u = 500 tickets

s = 50 tickets

We want to find the probability that the mean will be greater than 510. It is expressed as

P(x greater than 510) = 1 - P(x lesser than or equal to 510)

For x = 510

z = (510 - 500)/50 = 0.2

Looking at the normal distribution table, the probability corresponding to the z score is 0.9773

P(x greater than 510) = 1 - 0.9773 = 0.0227

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3 years ago

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ABCD is a parallelogram. If m

ikadub [295]

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3 years ago

I will mark you BRAINLIST please help with all 3 and show work on a paper please I need this thank you !!

o-na [289]

I hope you can understand this :)

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3 years ago

Olivia is making 15 bead bracelets for her friends what is the proportional relationship?

yanalaym [24]

5 minutes per bracelet.

because 15/3 = 5 mins.

To find the constant of proportionality in minutes per bracelet, divide the total time by the number of bracelets:

constant of proportionality =15/3= 5 minutes.

Now, we're going to consider an example of a proportional relationship in our everyday life: When we put gas in our car, there is a relationship between the number of gallons of fuel that we put in the tank and the amount of money we will have to pay. In other words, the more gas we put in, the more money we'll pay.

Learn more about the proportional relationships at

brainly.com/question/12242745

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2 years ago