How does 2^(1-i) expand to 2cos(ln 2)−2i sin(ln 2)?

[btbam91]

Hey guys, I'm having trouble on understanding how:

2^(1-i) expands to 2cos(ln2)-2i(sin(ln2))

Thanks in advance!

 

[logarithmic]

Hey guys, I'm having trouble on understanding how:

2^(1-i) expands to 2cos(ln2)-2i(sin(ln2))

Thanks in advance!

2^(1-i) = exp(ln[2^(1-i)]), go from there.