Taking the principal square root of a complex number

[burgerandcheese]

When solving (2 - i)z² + (4 + 3i)z + (-1 + 3i) = 0 by using the quadratic formula,

z = [-(4 + 3i) ± √(3 - 4i)] / (2(2 - i))

But what does √(3 - 4i) mean? Since when we take the principal square root of a real number we get its positive square root, but complex numbers don't have positive or negative square roots. The square roots of 3 - 4i are ±(2 - i)

How come the next line would be z = [-(4 + 3i) ± (2 - i) ] / (2(2 - i))

 

[ksdhart2]

Strictly speaking, it's true that under the standard definition of principal square root, "the unique nonnegative square root of a nonnegative real number," there's no such thing as the principal square root of a complex number because complex numbers are neither negative nor positive. However, many people extend this definition a little bit by making the arbitrary choice to define the principal square root of a complex number as the one written without a "minus sign." That is to say, we would define \(\sqrt{-1} = i\), \(\sqrt{-4} = 2i\), \(\sqrt{-17} = i\sqrt{17}\), etc.

 

[Dr.Peterson]

In this context, you don't need to choose either root as the principal square root, as you are going to use both anyway (±).

The concept was discussed in a recent thread, where we pointed out that the definition used for real radicands doesn't apply here, as there are no "positive" complex numbers; in cases like yours, in fact, both roots (2 - i and -2 + i) have a negative sign somewhere, so the convention mentioned by ksdhart2 (applicable to pure imaginary numbers) can't be used either.

We can define a principal root as we wish; one common choice is the root with the smallest positive argument (angle from the positive x-axis), which amounts to requiring the imaginary part to be positive. In that case, you would take -2 + i.

But since, as I said, it doesn't matter, they evidently just took the first root they thought of, which is common practice. Did they indicate how they got the root?

 

[burgerandcheese]

Thank you for your replies.

Did they indicate how they got the root?

Of course. They let (a^2 - b^2) + 2abi = 3 - 4i.

By the way, would it be wrong to put the equivalence symbol instead of the equal sign there?

(a^2 - b^2) + 2abi ≡ 3 - 4i.

 

[Dr.Peterson]

Of course. They let (a^2 - b^2) + 2abi = 3 - 4i.

There are several ways to do it, and in fact more than one way to proceed from here. One way or another, they presumably got a = ±2, which led to the two solutions ±(2 - i). They may or may not have at any point isolated 2 - i as "the" root; in any case, it doesn't matter whether they wrote ±(2 - i) or ±(-2 + i), as they are the same thing.

By the way, would it be wrong to put the equivalence symbol instead of the equal sign there?

(a^2 - b^2) + 2abi ≡ 3 - 4i.

The equivalence symbol would mean "for all a and b", which is not true. So that would be wrong.

I imagine you may be thinking of a similar situation where we want two polynomials to be the same (≡), so we set coefficients of variables equal. The equivalent here (no pun intended) would be that 1 and i take the place of x and y, setting their "coefficients", the real and imaginary parts, equal; since they are not variables, it wouldn't be appropriate.

 

[pka]

When solving (2 - i)z² + (4 + 3i)z + (-1 + 3i) = 0 by using the quadratic formula,

Reasonable mathematicians can disagree. I am sure there a others who do not agree with me here. But I would advise against applying the radical sign to any complex number that is not a non-negative real number.
So how do we use the quadratic formula with \(\displaystyle (2 - i)z^2 + (4 + 3i)z + (-1 + 3i) = 0\)
Will \(\displaystyle a=(2-i),~b=(4+3i),~\&~c=(-1+3i)\) from which we get the discriminant \(\displaystyle \Delta=b^2-4ac=3-4i\), see here
We need the two square roots of \(\displaystyle 3-4i=5\exp\left(\theta i\right)\) where \(\displaystyle \theta=\arctan\left(\frac{-4}{3}\right)\) see here
One square root of \(\displaystyle \Delta\) (the principal one I say) is \(\displaystyle \sqrt5\exp\left(\frac{\theta i}{2}\right)\). The other just add \(\displaystyle \pi\) to \(\displaystyle \theta\).