# No distinction between positive and negative for imag nums?

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?

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.

jbriggs444
Homework Helper
2019 Award
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.
But neither i nor -i are positive. There is no algebraic test that can distinguish between ##i=\sqrt{-1}## and ##j=-\sqrt{-1}##

pwsnafu
This is clearly untrue. Just consider the example ##w = a + 0i## and ##z = -(a + 0i)##. Clearly there is a distinction.
The quote says "nonreal numbers". You need to give examples from ##\mathbb{C}\setminus\mathbb{R}##.

Svein
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?
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...

ehild
Homework Helper
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.

mfb