Question

Let \( f: \mathbb{R}^2 \to \mathbb{R}^2 \) be defined as
\[ f(x, y) = (e^x \cos y,\ e^x \sin y) \]
Which one of the following is TRUE?

• A. \( f \) is one-to-one.
• B. The Jacobian of \( f \) is negative.
• C. \( f \) is locally invertible.
• D. \( f \) is invertible on \( \mathbb{R}^2 \).

Hint:

Compute the Jacobian determinant of f, then compare local invertibility with global injectivity using the periodicity of sine and cosine in y.

Answer 1

The Correct Option is C

Look at this map through complex numbers instead of raw partial derivatives. Write $z = x + iy$. Then $e^z = e^x(\cos y + i \sin y) = e^x \cos y + i\, e^x \sin y$, so the given $f(x,y) = (e^x\cos y, e^x \sin y)$ is exactly the real and imaginary parts of the complex exponential $e^z$.

1. (A) f is one-to-one: Since $e^z$ has period $2\pi i$ in $z$, the points $z=0$ and $z = 2\pi i$, that is $(x,y)=(0,0)$ and $(x,y)=(0,2\pi)$, both give $e^z = 1$. Two distinct inputs give the same output, so $f$ is not injective. This option is false.
2. (B) The Jacobian of f is negative: A holomorphic function like $e^z$ has Jacobian determinant equal to $|e^z|^2 = e^{2x}$, which is always strictly positive, never negative. This option is false.
3. (C) f is locally invertible: The complex derivative of $e^z$ is $e^z$ itself, and $e^z \neq 0$ for every $z$, so the map is a local diffeomorphism, meaning it is locally one-to-one with a smooth local inverse, at every point. This matches exactly what "locally invertible" means. This option is true.
4. (D) f is invertible on R2: Global invertibility needs global injectivity, which is already ruled out in option (A) using the $2\pi i$ periodicity. So $f$ cannot be invertible on all of $\mathbb{R}^2$. This option is false.

So the only correct statement is that $f$ is locally invertible everywhere, even though the periodicity of $e^z$ stops it from being globally one-to-one. This is a classic example showing that a nonzero Jacobian only guarantees a local inverse, not a global one.

Let's summarize:

• The Jacobian determinant of $f$ works out to $e^{2x}$, which never vanishes and is always positive.
• A nonzero Jacobian at a point gives local invertibility there by the Inverse Function Theorem, but says nothing about injectivity on the whole plane.
• The $2\pi$ periodicity in $y$, equivalently $2\pi i$ periodicity of $e^z$, is what breaks global injectivity, so options (A) and (D) both fail.

Hence the correct choice is that $f$ is locally invertible.

\[ \boxed{\text{(C) } f \text{ is locally invertible everywhere on } \mathbb{R}^2} \]