The harmonic conjugate of $\Im e^{z^2}$? It is obvious that $e^{z^2}$ is analytic, right? So the harmonic conjugate of $\Im e^{z^2}$ is $\Re e^{z^2}$, isnt' it?
However, the solutions manual I'm consulting gives the answer as $\Im (-ie^{z^2})$, which is not the same function, and I don't understand.
 A: For real-valued functions $u(x,y)$ and $v(x,y)$, we say that $v$ is a harmonic conjugate of $u$ if $u+iv$ is analytic. This is not a symmetric relation between $u$ and $v$. For example, $e^x\sin y$ is a harmonic conjugate of $e^x\cos y$ because $e^x\cos y+ie^x\sin y=e^z$ is analytic, but $e^x\cos y$ is not a harmonic conjugate of $e^x\sin y$ because $e^x\sin y+ie^x\cos y$ is not analytic (check the Cauchy-Riemann equations).
If $v$ is a harmonic conjugate of $u$, meaning that the function $f(z)=u+iv$ is analytic, then $-if(z)=v-iu$, being a constant multiple of $f(z)$, is also analytic; this shows that $-u$ is a harmonic conjugate of $v$, in other words, $-\Re f(z)$ is a harmonic conjugate of $\Im f(z)$.
To repeat one more time: although $\Im f(z)$ is a harmonic conjugate of $\Re f(z)$ (assuming $f(z)$ is analytic), it's $-\Re f(z)$, not $\Re f(z)$, that is a harmonic conjugate of $\Im f(z)$. This is true for any analytic function $f(z)$, including your problem function $f(z)=e^{z^2}$.
A: Let $w = a + bi$. Note that $\text{Im } (iw) = \text{Im } (-b + ai) = a = \text{Re } (w)$.
So yes, it is the same function up to a minus sign, which I don't know where it comes from - perhaps it comes from the manual's specific definition of a harmonic conjugate.
