# Thread: function values of complex functions

1. ## function values of complex functions

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!

2. Originally Posted by Hollysti
a. e^(4 + 2i)
This one's easy.

e^{4 + 2i} = e^4 * e^{2i}

= e^4 * (cos(2) + i*sin(2))

-Dan

3. Originally Posted by Hollysti
c. sin (4(pi) - (pi)i/2)
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

4. Originally Posted by Hollysti
b. Ln(5 - 2i)
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

5. Originally Posted by Hollysti
d. (1 + i)^(2i)
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