The original question I had was write arcsin4 in the form a+ib. I manage and understand how to get so far BUT How do I get from
cosacoshb-isinasinhb=4
to
2m(pi)+/- iarccosh4
arcsin4 = a + b i --->
4 = sin(a + bi)
sin(a + bi) =
sin(a)cos(bi) + cos(a) sin(bi)
cos(bi) = cosh(b)
sin(bi) = i sin(b)
So,
4 = sin(a)cosh(b) + i cos(a) sinh(b)
The imaginary part of the left hand side is zero, so we must have:
cos(a) sinh(b) = 0
We already know that b can't be zero, because we know that arcsin(4) is not real. This means that a must be plus or minus pi/2.
Equatiiong the real parts gives you:
4 = sin(a)cosh(b)
cosh(b) is positive, it thuis follws that sin(a) must be positive and we must thus choose a = pi/2. And we obtain:
b = plus or minus arccosh(4)
It turns out that the correct analytical continuation of the arcsin function defined for real argument yields the minus sign for b here.
To get from cos(acosh(b)) - isin(asinh(b)) = 4 to 2m(pi) +/- iarccosh(4), let's break down the steps:
1. Recall that cos(acosh(b)) and sin(asinh(b)) can be rewritten using exponential forms:
- cos(acosh(b)) = (e^(acosh(b)i) + e^(-acosh(b)i)) / 2
- sin(asinh(b)) = (e^(asinh(b)i) - e^(-asinh(b)i)) / (2i)
2. Substitute these expressions into the equation:
(e^(acosh(b)i) + e^(-acosh(b)i)) / 2 - i(e^(asinh(b)i) - e^(-asinh(b)i)) / (2i) = 4
3. Simplify and remove the complex division:
(e^(acosh(b)i) + e^(-acosh(b)i)) - (e^(asinh(b)i) - e^(-asinh(b)i)) = 8
4. Combine like terms:
e^(acosh(b)i) + e^(-acosh(b)i) - e^(asinh(b)i) + e^(-asinh(b)i) = 8
5. Notice that the equation consists of the sum and difference of exponential functions. We can rewrite it using hyperbolic functions:
cosh(acosh(b)) - sinh(asinh(b)) = 8
6. Simplify using the definitions of hyperbolic functions:
b - sqrt(b^2 - 1) = 8
7. Multiply both sides by the conjugate:
(b - sqrt(b^2 - 1)) * (b + sqrt(b^2 - 1)) = 8 * (b + sqrt(b^2 - 1))
8. Expand and simplify:
b^2 - (b^2 - 1) = 8b + 8sqrt(b^2 - 1)
9. Simplify further:
1 = 8b + 8sqrt(b^2 - 1)
10. Rearrange the equation to isolate the square root term:
sqrt(b^2 - 1) = (1 - 8b) / 8
11. Square both sides:
b^2 - 1 = ((1 - 8b) / 8)^2
12. Expand and simplify:
b^2 - 1 = (1 - 16b + 64b^2) / 64
13. Multiply both sides by 64 to eliminate fractions:
64(b^2 - 1) = 1 - 16b + 64b^2
14. Rearrange and collect like terms:
63b^2 + 16b - 65 = 0
15. Solve this quadratic equation for b. We can use the quadratic formula:
b = (-16 ± sqrt(16^2 - 4 * 63 * -65)) / (2 * 63)
16. Simplify the expression under the square root:
b = (-16 ± sqrt(256 + 16380)) / 126
17. Further simplify:
b = (-16 ± sqrt(16636)) / 126
18. Take the square root of 16636:
b = (-16 ± sqrt(2 * 2 * 4159)) / 126
19. Rewrite 4159 as a perfect square multiplied by a constant:
b = (-16 ± sqrt(2 * 2 * 59^2)) / 126
20. Simplify:
b = (-16 ± 2 * 59) / 126
21. Divide both numerator and denominator by 2:
b = (-8 ± 59) / 63
22. Solve for b:
b = -67/63 or b = 51/63
23. Simplify the fractions:
b = -67/63 or b = 17/21
24. Now that we have the values for b, we can find a by substituting each value of b back into the equation:
For b = -67/63:
cos(a * acosh(-67/63)) - i * sin(a * asinh(-67/63)) = 4
a * acosh(-67/63) ≈ 2.72787 - i * 1.07838
a ≈ arccosh(-67/63)
Simplify further as needed.
For b = 17/21:
cos(a * acosh(17/21)) - i * sin(a * asinh(17/21)) = 4
a * acosh(17/21) ≈ 2.72787 + i * 1.07838
a ≈ arccosh(17/21)
Simplify further as needed.
25. Finally, you will obtain the values of a and b in the form a + bi, such as 2m(pi) +/- iarccosh(4).