Powers Get Sloppy With Complex Numbers

Algebra Level 1

True or false :

\quad Let i = 1 i = \sqrt{-1} , then i 4 = 1 i^4=1 . Taking its fourth root gives i = 1 i = 1 .

True False

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3 solutions

The fourth root of 1 has many values.

Laurent Shorts
Apr 8, 2016

You could do the same without complex numbers:

Let a = 1 a=-1 , then a 2 = 1 a^2=1 . Taking its square root gives a = 1 a=1 .

The error is easy to spot: a n = 1 a = 1 a^n=1 \Leftarrow a=1 always, but a n = 1 a = 1 a^n=1 \Rightarrow a=1 only for real numbers and n n odd.


Note: b \sqrt{b} for b b a non-negative real number is the real positive number a a such that a 2 = b a^2=b and it's well defined because it is unique. But you cannot do the same with b b negative or complex. So you should not write 1 \sqrt{-1} . That leads to errors such as 1 = 1 1 1 = 1 1 1 = 1 \sqrt{-1}=\sqrt{\frac{1}{-1}} \Rightarrow \sqrt{-1}=\frac{1}{\sqrt{-1}} \Rightarrow -1=1 . You should prefer to define i i as a root of 1 -1 , that is, a number such that i 2 = 1 i^2=-1 .

Anish Harsha
Mar 28, 2016

Substituting i = 1 i=\sqrt{-1} in i 4 = 1 i^4=1 , then

( 1 ) 4 = 1 , 1 = 1 4 (\sqrt{-1})^4=1 \ , \ \sqrt{-1}= \sqrt[4]{1}

= 1 = 1 = \sqrt{-1} = 1

= 1 = 1 2 = -1 = 1^2

= 1 = 1 = -1 = 1

Thus, it would be invalid.

Your third step is wrong. 1 1 √-1 ≠ 1

akash patalwanshi - 5 years, 2 months ago

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Yeah, that's why its invalid.

Anish Harsha - 5 years, 2 months ago

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