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2 years ago

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Can someone give me the answer THE REAL ONE

1 answer:

ipn [44]2 years ago

3 0

The third one is the correct answer, i’m pretty sure! :))

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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

A plumber charges $75 to visit your house plus $50 per hour in labor costs to make any

Illusion [34]

Answer:

75+50h=A

Step-by-step explanation:

75 bucks plus 50 per hour equals the total cost

6 0

3 years ago

What is the proportion in the first question?

Igoryamba

If two triangles are similar, the correspondant sides are in proportion. So, the proportion is

$15\div 10 = 18 \div 12 = x \div 6$

4 0

2 years ago

Twenty seven minus 3/2 of a nuber x is not more than 36. What is the number?

Irina-Kira [14]

Answer:

Step-by-step explanation:

27 - 1.5x ≤ 36

54 - 3x ≤ 72

54 - 72 ≤3x

-18 ≤ 3x

-6 ≤ x

x ≥ -6

3 0

2 years ago

GEOMETRY!!!! K' (6,5) is the image of K after a reflection in the line y=2. what are the coordinates of K

frutty [35]

Given:

The coordinates of point K' are (6,5).

K' is the image of K after a reflection in the line y=2.

To find:

The coordinates of point K.

Solution:

Let the coordinates of point K are (a,b).

If a figure is reflected over the line y=2, then

$(x,y)\to (x,2(2)-y)$

$(x,y)\to (x,4-y)$

Using this formula, the coordinates of image of K are

$K(a,b)\to K'(a,4-b)$

The coordinates of point K' are (6,5).

$K'(a,4-b)=K'(6,5)$

On comparing both sides, we get

$a=6\text{ and }4-b=5$

$a=6\text{ and }-b=5-4$

$a=6\text{ and }-b=1$

$a=6\text{ and }b=-1$

Therefore, the coordinates of point K are (6,-1).

3 0

2 years ago