B No distinction between positive and negative for imag nums?

1. Aug 1, 2016

Mr Davis 97

I am solving for the square root of a complex number, namely: $\sqrt{5 -12i}$ Let's assume for brevity's sake that I already have the solution $3- 2i$. This is fine and well, but the book that I am reading goes on to say that "$- (3 -2i)$ is an equally valid solution, since there is no distinction between 'positive' and 'negative' for nonreal numbers." How is this so? Why do both of these solution work for the root of a complex number, while if I were calculating $\sqrt{5 - 12 \sqrt{2}}$ there would only be one solution, namely the positive one?

2. Aug 1, 2016

axmls

This is clearly untrue. Just consider the example $w = a + 0i$ and $z = -(a + 0i)$. Clearly there is a distinction.

Now, for example, in the case of the reals, there is a difference between finding the square root of a number and solving for the roots of a polynomial. If I wrote "what is $\sqrt{4}$?" then the answer would be $2$, but if I asked for the roots of $x^2 - 4 = 0 \Rightarrow x^2 = 4$, then there are clearly two solutions, $-2$ and $2$.

But this is just common practice. When people write the square root symbol with no context, they mean the positive solution; however, clearly the negative solution squared will yield the term in the square root as well.

3. Aug 1, 2016

jbriggs444

But neither i nor -i are positive. There is no algebraic test that can distinguish between $i=\sqrt{-1}$ and $j=-\sqrt{-1}$

4. Aug 1, 2016

pwsnafu

The quote says "nonreal numbers". You need to give examples from $\mathbb{C}\setminus\mathbb{R}$.

5. Aug 2, 2016

Svein

When you are in the complex domain there are new rules and new ways to look at things. Example: When you say $5 -12i$ I immediately convert it to $r\cdot e^{i\phi}$ with r being 13 and φ being -1.17601. From there you can take the square root easily - (square root of r and φ/2). But in doing so, you forget that $e^{i\phi}$ and $e^{i(\phi+2\pi)}$ represent the same number! So, taking the square root you must include both the solution with φ/2 and the solution with φ/2+π. And, since $e^{i\pi}=-1$, you see why the other solution is included...

6. Aug 2, 2016

ehild

The n-th root of an imaginary number Z is not unique. It means all the numbers z1, z2 z3 ..... the n-th power of which is Z: (zk)n=Z.

You can find the second root of 5-12i as a+ib so as (a+ib)2=5-12i. This means real equations for both the real part and imaginary part:
a2-b2=5, 2ab=-12. Eliminating b and solving the quadratic equation for a2, only one solution is positive, a2=9, so the solutions are a = 3 and b = -2; and a = -3, b = 2.

As for the complex n-th root, the n-the roots of the unit 1 are cos((2π/n)k)+isin((2π/n)k) for k=0, 1, n-1, or, in Euler's notation e((2π/n)ki). So the second roots of 1 are cos(0)+i sin(0) = 1 and cos(π) +i sin(π) = -1.
The n-th roots of a general complex number $Z=|Z|e^{iΦ}$ are $\sqrt[n] {|Z|}e^{i(Φ/n + (2π/n) k }$ for k=0, 1, 2...n-1 where |Z|1/n is the real n-th root of |Z|.
Written in trigonometric form, $Z^{1/n}=\sqrt[n] {|Z|}(\cos(Φ/n + (2π/n) k)+i\sin(Φ/n + (2π/n) k)$.
This way the second roots of Z=5-12i are = √13 (cos(Φ/2+πk)+i sin( Φ/2+πk) ), where Φ=arctan(-12/5) cos(Φ)=5/13, sin(Φ)=-12/13, and k = 0, 1.