# Thread: Square rooting complex numbers

1. ## Square rooting complex numbers

If A=3-41, find sqrtA
if you let x=sqrtA, there is only one solution, which is 2-i (first degree polynomial)
However, if you use the method of squaring, so x^2 = A, there are 2 solutions. +/-(2-i) (second degree polynomial)
This also arises when you do, for example, x=sqrt4=2 but x^2=4 gives x=+/-2
for sqrt4 the answer is 2 and not -2 because you take the positive case. However, a complex number cannot be positive or negative, so what would be the correct answer to sqrtA. [the answer says +/-(2-i) :/]

2. ## Re: Square rooting complex numbers

The answer can be plus negative

z = 3 - 4i

x^2 - y^2 = 3

x^2 + y^2 = 5

2x^2 = 8, x = (+/-) 2

2y^2 = -2, y = (+/-) 1

Since Im(z) < 0, opp in sign

Therefore Answer is +/-(2-i)

3. ## Re: Square rooting complex numbers

Originally Posted by mathpie
If A=3-41, find sqrtA
if you let x=sqrtA, there is only one solution, which is 2-i (first degree polynomial)
However, if you use the method of squaring, so x^2 = A, there are 2 solutions. +/-(2-i) (second degree polynomial)
This also arises when you do, for example, x=sqrt4=2 but x^2=4 gives x=+/-2
for sqrt4 the answer is 2 and not -2 because you take the positive case. However, a complex number cannot be positive or negative, so what would be the correct answer to sqrtA. [the answer says +/-(2-i) :/]
$\noindent Every non-zero complex number will have exactly \textbf{two} square roots. So we can't talk about the'' square root of a general complex number z unless we make a choice and rule telling us which square root we want to talk about. This requires us to take a branch'' of the square root function, but discussion of branches is not part of the HSC syllabus. There is something called the \textbf{principal branch} of the square root function, which gives us the \textbf{principal square root} of a complex number z. The principal square root of z\in \mathbb{C} is essentially defined as the square root of z (out of its two square roots) whose principal argument lies in the interval -\frac{\pi}{2} < \theta \leq \frac{\pi}{2}. Using this definition, every complex number will have a unique, unambiguous principal square root.$

4. ## Re: Square rooting complex numbers

For HSC purposes you should probably give both square roots as the answer in general (as a +/- pair).

5. ## Re: Square rooting complex numbers

I like the method in Terry Lee's:

$3 - 4i = 4 - 4i -1 = 2^2 -2\times 2 \times i + i^2 = (2 - i)^2 \\ \\ \therefore \sqrt{3-4i} = \pm (2-i)$

6. ## Re: Square rooting complex numbers

Originally Posted by Drongoski
I like the method in Terry Lee's:

$3 - 4i = 4 - 4i -1 = 2^2 -2\times 2 \times i + i^2 = (2 - i)^2 \\ \\ \therefore \sqrt{3-4i} = \pm (2-i)$
Is this sufficient for, say q11 2 marker?

7. ## Re: Square rooting complex numbers

Originally Posted by pikachu975
Is this sufficient for, say q11 2 marker?
It's essentially done by inspection and showing that the inspected answer works, so I don't see why not. (It's using the well-known fact that complex numbers will have only two square roots, but this should be assume-able (and expected to be known) knowledge for 4U students.)

8. ## Re: Square rooting complex numbers

Originally Posted by Drongoski
I like the method in Terry Lee's:

$3 - 4i = 4 - 4i -1 = 2^2 -2\times 2 \times i + i^2 = (2 - i)^2 \\ \\ \therefore \sqrt{3-4i} = \pm (2-i)$
I liked his method too, but never used it as I thought it had too less working out with an explanation.

9. ## Re: Square rooting complex numbers

Originally Posted by si2136
I liked his method too, but never used it as I thought it had too less working out with an explanation.
I don't see why a longer working out method is more legitimate. In fact in mathematics, we are always looking for shorter and more elegant solutions.

In some questions where finding the square root of a complex number is but a small part of the overall problem, you will not want to go thru the rigmarole of the long-winded method.

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