We know that \(i=\sqrt { -1 } \)

But \(i={ i }^{ 4\times \frac { 1 }{ 4 } }\\ { ({ i }^{ 4 }) }^{ \frac { 1 }{ 4 } }\\ { 1 }^{ \frac { 1 }{ 4 } }\\ =1\)

So does that mean i=1? Please help

We know that \(i=\sqrt { -1 } \)

But \(i={ i }^{ 4\times \frac { 1 }{ 4 } }\\ { ({ i }^{ 4 }) }^{ \frac { 1 }{ 4 } }\\ { 1 }^{ \frac { 1 }{ 4 } }\\ =1\)

So does that mean i=1? Please help

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TopNewestHmm. 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. :-) – Sachin Vishwakarma · 1 year, 3 months ago

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– Satyajit Ghosh · 1 year, 3 months ago

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.Log in to reply

– Sachin Vishwakarma · 1 year, 3 months ago

Yup. That is root. (Although that is false)Log in to reply

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. – Margaret Zheng · 1 year, 3 months ago

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– Satyajit Ghosh · 1 year, 3 months ago

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?Log in to reply

– Kushagra Sahni · 1 year, 2 months ago

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?Log in to reply