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3 years ago

13

Write each vector in component form

1 answer:

meriva3 years ago

5 0

Answer to problem 1 is (9,9)
Answer to problem 2 is (0,-4)

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Explanation:

To find the component form, they basically want the translation rule. To find the translation rule, we subtract the coordinates tip - head

In general, if we have two vectors M and N such that
M = (a,b)
N = (c,d)
then
vector MN = (c-a, d-b)

Applying this to the vector CD gives us
vector CD = (7-(-2), 10-1) = (7+2, 10-1) = (9, 9)
Basically meaning that we shift point C nine units to the right and nine units up to get to point D

Similarly,
vector RS = (-4-(-4), -3-1) = (-4+4, -3-1) = (0,-4)
so we shift point R four units down to get to point S

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

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Answer:

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Step-by-step explanation:

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

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Then,

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So,

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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

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

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

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

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