evaluate complex square root

Question

$\sqrt{i^4}$ $=\sqrt{1}=1$ or $=(\sqrt{i})^4$ $=(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}})^4$ $=\exp(4\ln\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$ $=\exp(4i\frac{\pi}{4})=\exp(\pi)i=-1$ That is in case of the principle argument of the power function I think the second solution is the correct because i used the definition of the power function Does it correct or not ?

Answer 1

The square rooth is not a well defined function in the complex numbers field.

$u=\sqrt{z}$ is a so called multi-valued function with two different values that are the two complex numbers $u_1,u_2$ such that $u_1^2=u_2^2=z$.

In your case , since $i^4=1$ we have the two values $\sqrt{1}=\pm 1$.

If we use the symbol $\sqrt{\cdot}$ for the principal value of the complex square root, than, with the usual definition of the principal value, we have, in this case: $$ \sqrt{i^4}=\sqrt{e^{i2\pi}}=e^{i\pi}=-1 $$

Note that this is different from the definition of the square root in the filed of real numbers, were $y=\sqrt{x}$ is the the positive number $y$ such that $y^2=x$

Answer 2

The convention is that when $0\leq x\in \Bbb R$ is that $0\leq \sqrt x\in \Bbb R.$ In particular $y=\sqrt x$ is an equation when $0\leq x\in \Bbb R.$ But $y=\sqrt x$ for non-real $x$ or for $0>x\in \Bbb R$ should not be taken literally as an equation. It only means that $x^2=y,$ and there are $2$ possible values $x$ might be.

Failure to follow this will result in apparent paradoxes.

Answer 3

Whenever we are talking about a one-to-many mapping, we need to specify what we mean when we perform operations on sets under such a mapping.

Because the mapping $$f : \mathbb C \to \mathbb C, \quad f(z) = z^2$$ satisfies $f(-z) = f(z)$ for all $z$, that is to say, every nonzero point in the image of $f$ has two distinct preimages in $\mathbb C$, the inverse mapping $$f^{-1} : \mathbb C \to \mathbb C, \quad f^{-1}(z) = z^{1/2}$$ is a one-to-two mapping: each nonzero $z \in \mathbb C$ will admit two images under $f$, and because, unlike the reals, $\mathbb C$ is not a totally ordered field, it is not necessarily obvious which of these is the image that is "intended."

Thus, if wish to be general and think of complex-valued mappings as operations acting on subsets of $\mathbb C$, then it is natural to conclude that the mapping $$g : \mathbb C \to \mathbb C, \quad g(z) = (z^4)^{1/2}$$ would imply that $$g(i) = (i^4)^{1/2} = (1)^{1/2} = \{-1, 1\}.$$ This has nothing to do with $i$ or complex numbers per se but rather the fact that $1$ has two square roots, even in the reals. Put simply, the meaning of $(1)^{1/2}$ is, "what is the set of all complex numbers such that their square equals $1$?" And the answer to this is of course $\{-1, 1\}$. Both are equally legitimate choices unless you impose some additional condition.