Complex Numbers Watch

qazwsxedc
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#1
Report Thread starter 9 years ago
#1
Find arccos4 in the form a + ib.

After giving up and looking at the answer, it gave:

2m\pi \pm iarcosh4

How did they derive that?
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Glutamic Acid
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#2
Report 9 years ago
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arccos(4) = a + ib so 4 = cos(a + ib); expand using compound angle formula, use the hyperbolic identities for cos(ib) and sin(ib) and equate real and imaginary parts.
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qazwsxedc
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#3
Report Thread starter 9 years ago
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(Original post by Glutamic Acid)
arccos(4) = a + ib so 4 = cos(a + ib); expand using compound angle formula, use the hyperbolic identities for cos(ib) and sin(ib) and equate real and imaginary parts.
Works like a charm - thank's GA
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qazwsxedc
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#4
Report Thread starter 9 years ago
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The second part of the question is tough:

Express arcsin2 in terms of a + ib

So: sin (a + ib) = 2

sin(a)cos(ib) + cos(a)sin(ib) = 2

sin(a)cosh(b) + icos(a)sinh(b) = 2

comparing real & imaginary:

(*)sin(a)cosh(b) = 2
(**)cos(a)sinh(b) = 0

Considering (**)
the only way that sinhb can form 0 is if b=0 which will mess up the imaginary part => cosa = 0 and so only
a = \frac{\pi}{2} , \frac{3\pi}{2}, \frac{5\pi}{2} etc... = \frac{\pi}{2} + \pi m
we must wait from deductions from (*) to fine-tune our values of 'a'.

Considering (*)
sin(a)cosh(b) = 2

a cannot be the -1's which sina produces =>

a = \frac{\pi}{2} , \frac{5\pi}{2}, \frac{9\pi}{2}  etc... = \frac{\pi}{2} + 2\pi m

b is simply arcosh2 [which is the correct answer]

However, my answer for 'a' is wrong; it should be:

(4m + 1) \frac{\pi}{2}
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ghostwalker
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#5
Report 9 years ago
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Your answer for a is correct, just take pi/2 out as a factor and you'll see.
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qazwsxedc
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#6
Report Thread starter 9 years ago
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(Original post by ghostwalker)
Your answer for a is correct, just take pi/2 out as a factor and you'll see.
Darn it :redface:

Thanks ghostwalker
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