If $i^4 = 1$ then isn't $i = 1$

Question

The imaginary unit "$i$" is equal to the square root of $-1$ by definition.

If we take $i$ to the forth power then we get $1$

But if $i^4 = 1$ then solving for i we get 1 instead of $\sqrt{-1}$

Can someone explain why this is to me?

Answer 1

A forth degree equation may have up to 4 different solutions. In this case the solutions are: $1,-1,i,-i$.

Answer 2

The entire concept of the square root (and higher roots) is problematic. Take the equation

$$(-3)^2 = 9$$

We could "solve" this to get $-3 = 3$. Clearly this is nonsense. For any real or complex number $z$, there are precisely $n$ numbers $x$ that have the property

$$x^n = z$$

This can be interpreted visually by using euler's formula, which essentially says that multiplying complex numbers is the same as rotating them on the complex plane. For more, see https://betterexplained.com/articles/understanding-why-complex-multiplication-works/.

Getting back to your original question, there are precisely four numbers $1$, $-1$, $i$, and $-i$ whose fourth powers are 1. This fact certainly does not mean $i =1$

Answer 3

What you are doing is starting with $i^4 = 1$ and then you take the fourth root of both sides to conclude that $(i^4)^{1/4}=1^{1/4}$ implies $i=1$. This is not correct.

If two things become equal when raised to the same power, it does not imply those two things were equal to begin with. I believe you may already be familiar with this in the context of real numbers. For example, $(-3)^2=3^2$ but $-3 \ne 3$, etc. This rule still applies even if we throw complex numbers into the mix.

More slightly related details/info:

When you just evaluate $1^{1/4}$ then it would be correct to say the answer is 1, because typically we're looking for the principal root unless otherwise specified. Just like how we customarily say $\sqrt{16}$ is $4$ and not $-4$.

This is different from solving the equation $x^4 =1$. The fundamental theorem of algebra tells us this equation has four complex roots. Specifically, they are $x=1,-1,i,-i$. Just like how the equation $x^2=16$ has the two solutions $x=4,-4.$

Answer 4

$i^4 = 1$ and $(-i)^4=1$ and $1^4=1$ and $(-1)^4 =2$.

This is no more a contradiction than $(-3)^2=9$ and $3^2=9$.

Remember if $x^2 = c $ you can NOT conclude $x = \sqrt {c} $ . You can only conclude $x = \sqrt {c} $ or $x = -\sqrt {c}$.

It's similar with 4th roots. If $x^4 = c $ then there are four possible values for $x $. $x $ can equal $\sqrt [4]{c} $ or $-\sqrt[4]{c} $ or $i\sqrt[4]{c}$ or $-i\sqrt[4]{c} $.

It get's very interesting when trying to solve $x^6 = c$. There are six possible values. $x = \pm\sqrt [6]{c} $. or $x=\pm\sqrt [6]{c}/2 \pm \sqrt {3}\sqrt [6]{c}i/2$.

There's actually a trick to these that you will learn.