How does the process of simplifying imaginary numbers actually work?

Question

Sorry in advance if this is a really stupid question

In class I've been told that $$\sqrt{-25} = 5j $$ Converting $\sqrt{-25} $ into $5j$ is straightforward for me, but I don't understand how it works

Doesn't the property of $\sqrt{xy} = \sqrt{x}\sqrt{y} $ only hold true for positive real number values of x and y, where $i^2$ is defined to be negative 1?

In the case of $$\sqrt{-25} = 5j $$ Would we treat j= $\sqrt{-1}$ as a positive real number in order to "break" the root using the elementary algebra associated with roots? I don't really understand how this works algebraically.

Answer 1

There's no such thing as a stupid question!

As other people have pointed to, the square-root function isn't really well-defined for all complex numbers. Fix some complex number $z\neq 0$. We can try to solve the quadratic equation $w^2=z$ (that's what a square root seeks to do) and we will find two answers $\pm w$. Which one we call $+w$ and which one we call $-w$ is purely convention. The fact that these answers are negatives of each other comes from $(-w)^2=w^2$, which is really just a dressed-up version of $(-1)^2=1$. The fact that there are always two answers comes from the "Fundamental Theorem of Algebra."

If $z$ is a positive real number, then its square roots are of the form $\pm w$, where $w>0$. Of course, choosing $w>0$ was arbitrary, but it allows us to concretely write $\sqrt{z}$ as another way to denote this positive square-root $w$.

If $z$ is a negative real number, then you can set a similar convention and take its square-roots to be $\pm iw$, where $w>0$. With this, you can write $\sqrt{z}=iw$, although this notation is not always used. This is the same as defining $\sqrt{z}=i\sqrt{-z}$, using the existing convention for positive square roots.

As you've noted, even the very symbol $i$ is only defined up to a factor $\pm 1$, because $i^2=(-i)^2=-1$. This can be phrased by saying that complex conjugation $\overline{a+bi}=a-bi$ is an isomorphism of fields. This is probably getting beyond the scope of what you wanted to know (though I'm happy to answer further questions).

Answer 2

Doesn't the property of $\sqrt{xy} = \sqrt{x}\sqrt{y} $ only hold true for positive real number values of $x$ and $y$?

No, not "only"! The principal square root function on the complex numbers, $\sqrt{\cdot}:\mathbb C\to\mathbb C$, satisfies this identity for a much wider range of values. See When does $\sqrt{wz}=\sqrt{w}\sqrt{z}$? and When does $\sqrt{a b} = \sqrt{a} \sqrt{b}$? for details.

For your purposes, it suffices to know that the identity still holds if at least one of $x$ and $y$ is positive. So you can conclude that $\sqrt{-25}=\sqrt{-1}\times\sqrt{25}$.

Answer 3

It's all a matter of convention. The square root symbol, $\sqrt z$, always refers to a number whose square is equal to $z$, and, except for $z=0$, there are always two choices. In order to think of $\sqrt z$ as a well defined function, it's necessary to specify which choice is made, and that's a question of convention.

There are two standard conventions, and you can sometimes get in trouble if you don't know which one is being used. In one convention the real part of the square root is always non-negative, and in the other the imaginary part is always non-negative; in both conventions, the square root of a positive real has postive real part while the square root of a negative real number has positive imaginary part.

The OP understands correctly that the "identity" $\sqrt{ab}=\sqrt a\sqrt b$ does not hold in general; forgetting that fact is the underlying cause of all kinds of paradoxical nonsense, such as "proofs" that $1=0$. However, the identity does hold for $a\in\mathbb{R^+}$ (and arbitrary $b$) under both of the standard conventions.

There's nothing aside from common sense to prevent one from inventing some fanciful, idiosyncratic convention such as saying the the square root of an odd negative integer has positive imaginary part while the square root of any other negative real has negative imaginary part. The identity $\sqrt{-25}=5i$ would still hold under such a convention, but we would have $\sqrt{-16}=-4i$ instead. Good luck, though, getting anyone to agree to use this convention.

Answer 4

$-25=25\cdot(-1)=25i^2$ so $\sqrt{-25}=5i$ because $(5i)^2=25i^2=-25$.

I have only used that $i^2=-1$. I haven't used the property $\sqrt{x\cdot y}=\sqrt{x}\cdot\sqrt{y}$.

Answer 5

Just another way of saying the same thing others have said:

You're rediscovering something you probably already learned in another context. $5 \cdot 5 = 25$, and $(-5)\cdot (-5) = 25$. So, if by $\sqrt{x} = k$ we mean $k$ is a number such that $k^2 =x$, then $\sqrt{25} = \pm 5$.

In the end this is just an ambiguity in the way that $\sqrt{\cdot}$ is defined. Some times it's defined in the algebraic way above. Other times, it's defined only for positive real numbers so that it can be a function. Other times still, it's made into a function by defining it on a certain subset of the complex plane.