compute the following: 1. 3^i 2. e^2log(-1)?

1. For complex numbers we use that

a^z = e^log(a^z) = e^(z*loga).

Then 3^i = e^(i log3) and you need to find the value of the complex logarithm

log z = log|z| + i Argz + 2k*pi*i, where k any integer

log 3 = log|3| + i Arg3 + 2k*pi*i = log3 + i*0 + 2k*pi*i = log3 +2k*pi*i.

Then

3^i = e^[ i*(log3 + 2k*pi*i)] = e^( i*log3 - 2k*pi) =[e^(i*log3)] * [e^(-2k*pi)]

According to Euler's formula e^(i*y) = cosy + i*siny, so

3^i = [cos(log3) + i*sin(log3)] *[e^(-2k*pi)] or

3^i = [e^(-2k*pi)] cos(log3) + i* [e^(-2k*pi)] sin(log3), k any integer. <== ANS

In case you use just principal value of the logarithm we get

3^i = e^(i*Log3) = e^[ i* (log|3| + i Arg3)] =

= cos(log3) + i* sin(log3) ≈ 0.455 + 0.891 i <=== principal value

2. e^2log(-1) = e^2*(log|-1| + i*Arg(-1) + 2k*pi*i) =

= e^2(log1 + i*pi + 2k*pi*i) = e^2(2k + 1)*i*pi = e^(4k + 2) * i*pi = 1.

1) 3^i = .4548324228 + 08905770417i

2) e^2log(-1) = -.915985097 + .4012122906i

1. 3^i not sure about

2. domain error on calculator... says... no. if it was equal to 1... then the calculator should yield an answer of 1... it doesn't.

also order of operations says... do the argument first.