Question about the solution of this system of equations

[DottZakapa]

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 ?

 

[Mark44]

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 $2\theta= -\theta-2k\pi , k\in \mathbb Z$

because it should be positive, right ?

 

[DottZakapa]

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]

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.

 

[LuccaP4]

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.