This problem says:
Showing the steps, not just giving the final answer, find the values of:
a. e^(4 + 2i)
b. Ln(5 - 2i)
c. sin (4(pi) - (pi)i/2)
d. (1 + i)^(2i)
This stuff will be on my final on Tuesday, and I don't really get how to do it at all. If someone would help me with these, that would be great. Please show me how you worked the problem so that I will be able to understand this stuff for the test. Thanks!
This one is easy if you recall that even for the complex extension of the sine function we still have:
sin(A + B) = sin(a)cos(B) + sin(B)cos(A)
So:
sin(4(pi) - i*(pi)/2) = sin(4(pi))cos(i*(pi)/2) - sin(i*(pi)/2)cos(4(pi))
= -sin(i*(pi)/2)
Now
sin(x) = (1/(2i))*(e^{ix} - e^{-ix})
So
sin(i*(pi)/2) = (1/(2i))*(e^{i*i*(pi)/2} - e^{-i*i*(pi)/2})
= (1/(2i))*(e^{-(pi)/2} - e^{(pi)/2})
= (1/i)*-sinh((pi)/2) = i*sinh((pi)/2)
-Dan
The function is "ln" by the way, not "Ln."
ln(z) = ln|z| + i*arg(z)
So
ln(5 - 2i) = ln|5 - 2i| + i*arg(5 - 2i)
Let's convert 5 - 2i to complex polar form:
5 - 2i = re^{it} (t being shorthand for "theta.")
r = sqrt{5^2 + (-2)^2} = sqrt{29}
t = atn{-2/5} = -atn{2/5}
So
5 - 2i = sqrt{29}*e^{-i*atn(2/5)}
Thus
ln|5 - 2i| = ln(sqrt{29}) = (1/2)*ln(29)
arg(5 - 2i) = -atn(2/5)
So
ln(5 - 2i) = (1/2)*ln(29) - i*atn(2/5)
-Dan
Let x = (1 + i)^{2i}
Then
ln(x) = (2i)*ln(1 + i)
ln(1 + i) = ln|1 + i| + i*arg(1 + i)
= (1/2)*ln(2) + i*atn(1) = (1/2)*ln(2) + i*(pi)/4
So
ln(x) = (2i) * ((1/2)ln(2) + i*(pi)/4)
= i*ln(2) - (pi)/2
= -(pi)/2 + i*ln(2)
Thus
x = e^{-(pi)/2 + i*ln(2)} = [e^{-(pi)/2}] * [e^{i*ln(2)}]
= [e^{-(pi)/2}] * [cos(ln(2)) + i*sin(ln(2))]
= [e^{-(pi)/2}]*[cos(ln(2)) + i*sin(ln(2))]
-Dan