Question

Euler's formula
\[2^{1-i}=2cos(ln2)-2isin(ln2)\]
I don't understand how it is instead of =2cos(2) -2isin(2)
Please, explain me

Answer 1

hmm, I haven't done these ones

Answer 2

I got it. \[\huge2^{1-i}=e^{ln2^{1-i}}\\\huge =e^{ln2-iln2}\]
It forms the form of complex number in Euler's formula
\[\huge e^{(ln2-iln2}=e^{ln2}[(cos (ln2)-isin (ln2)]\]
=\[\huge 2(cos(ln2)-isin(ln2))\]

Answer 3

laaalalalala life is beautiful!!!!