Question about the solution of this system of equations

Homework Statement:
complex equations
Relevant Equations:
complex numbers
hi
given such system of equations

##
\begin{cases}
\rho^2 = 2 \rho \\
2\theta= -\theta+2k\pi , k\in \mathbb Z \\
\end{cases}
##
in the solution of the professor the system is solved is solved as follows.
##
\begin{cases}
\rho=0 , \rho=2 \\
\theta= -\frac 2 3 k\pi , k = 0,1,2 \\
\end{cases}
##
isn't that minus in front of 2/3 an error?
because it should be positive, right ?

Last edited by a moderator:

Mark44
Mentor
Homework Statement:: complex equations
Relevant Equations:: complex numbers

hi
given such system of equations

##
\begin{cases}
\rho^2 = 2 \rho \\
2\theta= -\theta+2k\pi , k\in \mathbb Z \\
\end{cases}
##
in the solution of the professor the system is solved is solved as follows.
##
\begin{cases}
\rho=0 , \rho=2 \\
\theta= -\frac 2 3 k\pi , k = 0,1,2 \\
\end{cases}
##
isn't that minus in front of 2/3 an error?
It might have been an oversight, but it isn't incorrect.
The equation you wrote could just as well have been written as ##2\theta= -\theta-2k\pi , k\in \mathbb Z ##
DottZakapa said:
because it should be positive, right ?

It might have been an oversight, but it isn't incorrect.
The equation you wrote could just as well have been written as ##2\theta= -\theta-2k\pi , k\in \mathbb Z ##

if you solve the first system, from where does that minus in front of ##-2k\pi ## comes from?

Mark44
Mentor
if you solve the first system, from where does that minus in front of ##-2k\pi ## comes from?
Look at it this way:
##2\theta= -\theta+2k\pi , k\in \mathbb Z##
represents exactly the same set of numbers as
##2\theta= -\theta - 2k\pi , k\in \mathbb Z##
As a slightly different example, if ##\sin(\theta) = \frac 1 2##, then all solutions are given by
##\theta = \frac \pi 6 + 2k\pi, k \in \mathbb Z##
The same set of values for ##\theta## could just as well been written as ##\theta = \frac \pi 6 - 2k\pi, k \in \mathbb Z##.

Like I said, it might have been an oversight on the part of your instructor, but it doesn't make any difference in the resulting values.

As @Mark44 says, its the same expression. An integer ##k## could be positive or negative, so if you take ##2k## or ##-2k##, both determines the same set of integers, which is positive and negative even integers.

##k=\left\lbrace...,-3,-2,-1,0,1,2,3,...\right\rbrace ##
##2k=\left\lbrace...,-6,-4,-2,0,2,4,6,...\right\rbrace ##
##-2k=\left\lbrace...,6,4,2,0,-2,-4,-6,...\right\rbrace ##

They are both the same set (the order in which you write the elements of a set doesn't matter).
I don't know the context, but sometimes signs are chosen in order to simplify some expressions, or make it easier to realize something.