@Omair Siddique We know that e^ln(x)=x (I used e because I can't type logs with bases, but it works for any number, not just e), so we can rewrite the equation as the following:
2^i=(e^ln(2))^i=e^(i*ln(2))
We can use Euler's formula now, to determine the actual value. Euler's formula says that e^(ix)=cos(x)+i*sin(x).
@Omair Siddique For these questions, always transform the base in the form of e^ln(x). This simple thought can be used for questions like i^i and (a+bi)^(c+di).
cos(ln2)+isin(ln2)
Please explain the procedure
@Omair Siddique We know that e^ln(x)=x (I used e because I can't type logs with bases, but it works for any number, not just e), so we can rewrite the equation as the following:
2^i=(e^ln(2))^i=e^(i*ln(2))
We can use Euler's formula now, to determine the actual value. Euler's formula says that e^(ix)=cos(x)+i*sin(x).
e^(i*ln(2))=cos(ln(2))+i*sin(ln(2))=0.769+0.638i
Someone correct me if I'm wrong.
@Omair Siddique For these questions, always transform the base in the form of e^ln(x). This simple thought can be used for questions like i^i and (a+bi)^(c+di).