Quantum Mechanics

`\color{green}"The following steps explain How to handle a complex power"`

`x^{a+ib}=x^{a}.x^{ib}=x^{a}.(e^{log(x)})^{ib}because x=e^{log(x)}`

`=x^a.e^{ib.log(x)}=x^{a}[cos(b.log(x))+isin(b.log(x))]`  (by de Moivres theorem)

In the above result put a=0 and b=1 results in

`x^i=[cos(log(x))+isin(log(x))]`

(1)

(2)


(3)

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