Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f(r,\theta)=(r\cos\theta ,r\sin\theta)$ for $(r,\theta)$ $\in \mathbb R^2$ with $r\ne0$. Pick out the true statements?

1.$f$ is one-one on {$(r,\theta)$$\in\mathbb R^2$:$r\ne0$}

2.for any $(r,\theta)$$\in\mathbb R^2$ with $r\ne0$, $f$ is one-one 0n a neighborhood of $(r.\theta)$

I think Statement 1. is false since $r\ne0$, $f(r,2\pi)$=$f(r,4\pi)$ does not imply $(r,2\pi)$=$(r,4\pi)$

similar reason for 2

Am I right

share|cite|improve this question
up vote 1 down vote accepted

Your reasoning for (1) is fine, but (2) is actually true.

Let $\langle r,\theta\rangle$ be a point at which $r\ne 0$. Let $$N=\{\langle r_1,\theta_1\rangle:|r-r_1|<|r|\text{ and }|\theta-\theta_1|<\pi\}\;;$$ $N$ is a neighborhood of $\langle r,\theta\rangle$. Try to show that $f$ is one-to-one on $N$; I’ve complete the solution below, but I’ve left it spoiler-protected to give you a chance to try it your own. (Mouse-over to see it.)

Suppose that $\langle r_1,\theta_1\rangle\in N$ and $f(\langle r,\theta\rangle)=f(\langle r_1,\theta_1\rangle)$. Then $r\cos\theta=r_1\cos\theta_1$, and $r\sin\theta=r_1\sin\theta_1$. Let $a=\frac{r_1}r\ne 0$; then $\cos\theta_1=a\cos\theta$, and $\sin\theta_1=a\sin\theta$. At least one of $\sin\theta$ and $\cos\theta$ is non-zero; suppose that $\cos\theta\ne 0$. Then we can divide the equation $\sin\theta_1=a\sin\theta$ by $\cos\theta_1=a\cos\theta$ to get $\tan\theta_1=\tan\theta$. Since $|\theta-\theta_1|<\pi$, this implies that $\theta=\theta_1$. But then $\cos\theta=\cos\theta_1=a\cos\theta$, so $a=1$, and $r_1=r$. Thus, $\langle r_1,\theta_1\rangle=\langle r,\theta\rangle$, and $f$ is one-to-one on $N$. The argument if $\sin\theta\ne0$ is very similar.

share|cite|improve this answer

You are right. Intuitively, if you had a finite (and non-zero) rod on a plane and you moved it in some angular displacement $\theta$, there could be multiple ways of achieving same Cartesian position (x,y), by just moving in complete circle(s) which is adding multiple of $2\pi$ as you suggested. However if you had to move very slightly in any angular position, you would always get a unique Cartesian position. Hence 2 is correct and 1 is not.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.