I think 1 (one)e^(pi.i)+1=0 then(e^(pi.i)+1)^0 =1 and 0^0=1

No, you cannot say that.(e^(pi.i)+1)^0 is 0^0 which is still undefined

I think 1 (one)

ReplyDeletee^(pi.i)+1=0 then

(e^(pi.i)+1)^0 =1 and 0^0=1

No, you cannot say that.

ReplyDelete(e^(pi.i)+1)^0 is 0^0 which is still undefined