Here is what I think: 1^0.25=1/1^4. just as x^2=k(k is any constant) has 2 solutions (whether it is within the range of real number or not), x^4=k should have 4 solutions. Think that you are solving an equation 1/x^4 = 1. i and 1 are both solutions (two out of the four) of the equation 1/x^4=1.

I totally agree that it has 4 solutions, but if we open x then it will have 1 as a solution of i. Although there maybe 3 more but 1 is also its solution right?

Hmm. Good. What if we write it as $(1^ \frac {1}{2} )^ \frac{1}{2}$ , then square root of 1 is -1. Then you will get the desired result. Your working is a result of false root. :-)

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TopNewestHere is what I think: 1^0.25=1/1^4. just as x^2=k(k is any constant) has 2 solutions (whether it is within the range of real number or not), x^4=k should have 4 solutions. Think that you are solving an equation 1/x^4 = 1. i and 1 are both solutions (two out of the four) of the equation 1/x^4=1.

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I totally agree that it has 4 solutions, but if we open x then it will have 1 as a solution of i. Although there maybe 3 more but 1 is also its solution right?

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i can also be written as i^n×1/n and then i will have n roots. So now can we say that i has n values?

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Hmm. Good. What if we write it as $(1^ \frac {1}{2} )^ \frac{1}{2}$ , then square root of 1 is -1. Then you will get the desired result. Your working is a result of false root. :-)

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It's what Margaret was talking about but even if there is a false root this means 1 is a root of i irrespective of it being false or true.

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Yup. That is root. (Although that is false)

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