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2 years ago

Whose responsibility is it to call out the government and point out flaws

Mathematics

1 answer:

Jlenok [28]2 years ago

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Adults, 18 and older

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Find e^cos(2+3i) as a complex number expressed in Cartesian form.

ozzi

Answer:

The complex number $e^{\cos(2+31)} = \exp(\cos(2+3i))$ has Cartesian form

$\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2)$ .

Step-by-step explanation:

First, we need to recall the definition of $\cos z$ when $z$ is a complex number:

$\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}$ .

Then,

$\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}$ . (I)

Now, recall the definition of the complex exponential:

$e^{z}=e^{x+iy} = e^x(\cos y +i\sin y)$ .

So,

$e^{2i-3} = e^{-3}(\cos 2+i\sin 2)$

$e^{-2i+3} = e^{3}(\cos 2-i\sin 2)$ (we use that $\sin(-y)=-\sin y)$ .

Thus,

$e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)$

Now we group conveniently in the above expression:

$e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2$ .

Now, substituting this equality in (I) we get

$\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2$ .

Thus,

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)$

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right]$ .

5 0

2 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

mrs_skeptik [129]

2(2x − 1) > 6 or x + 3 ≤ −6

2(2x − 1) > 6

4x - 2 > 6

4x > 8

x > 2

x + 3 ≤ −6

x ≤ - 9

Solution: x ≤ - 9 or x > 2

(- ∞ , - 9] or (2 , + ∞)

Answer is the first one

(- ∞ , - 9] or (2 , + ∞)

8 0

2 years ago

Please solve the question

Margarita [4]

U should do like this... But sry I don't know third part

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2 years ago

250 is 1.5cm. How thick is 400 pages

svetlana [45]

Answer: 2.4 cm

Step-by-step explanation:

1.5 divided by 250 gives that each page is .006cm. 400 times .006 is 2.4cm

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2 years ago

What is the error in this problem

Vera_Pavlovna [14]

Step-by-step explanation:

The adjacent should be 16, 39 is the hypotenuse

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2 years ago