my question is related one topic ,which i would like to clarify using your help.so please help me to clarify this topic.problem is following(actually it is just my interest to clarify it)
in generally we can represent any real number as special case of complex number,or more corrected this case is when imaginary part is zero,like


or if we compare form $a+b*i$ --> we get $a=5$ and $b=0$

but because rules which works in real number domain ,does not work in complex domain like $\log(z_1)+\log(z_2)=\log(z_1*z_2)$ does not always works unless special case and also $\sqrt{-1}*\sqrt{-1}$ does not equal to $\sqrt{-1*-1}$, could we represent real number in the following form


because we know that every number in power 0(except 0) is 1,is it also true for imaginary $i$ as well?i am just curious it is not any homework,so please help me to clarify it


1 Answer 1


You have to think about how the expression $z^a$ is defined for complex $z, a$. What you do is you pick any number $c$ such that $e^c = z$ (we write $c = \log z$), and then you put $z^a = e^{ca}$. Note that there are usually many possible values for $z^a$, since there are many possible values of $c$. In your case, though, no mather what number $c$ you pick for $e^c = i$, you'll have $i^0 = e^{c \cdot 0} = e^0 = 1$, so indeed $5 = 2 + 3 i^0$, because $i^0 = 1$.

  • $\begingroup$ thanks very much.i was not sure about this $\endgroup$ Mar 31, 2013 at 9:36
  • $\begingroup$ because complex numbers have some constraints rules then real numbers.i wanted to know this fact $\endgroup$ Mar 31, 2013 at 9:38

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