Why are the two equations equal?

In some book, we can read

$$g(t,r)=\iint_{|z|\leqslant r} (iu_x-u_y)\text{d}x\text{d}y=i\iint_{|z|\leqslant r}(u_x+iu_y)\text{d}x\text{d}y=i\int_0^r\!\!\!\!\int_0^{2\pi}(u_x+ỉu_y)\rho\text{d}\rho\text{d}\theta$$Set $y=r^2$, $\psi(t,y)=g(t,r^2)$; then $\psi_y=g_rr_y=g_r/2r$, so that using our equation, we find$$\psi_y=\frac{1}{2}\int_{|z|=r^2}(u_x+iu_y)\frac{\text{d}z}{z}$$

My question is: why $\psi_y=\frac{1}{2}\int_{|z|=r^2}(u_x+i u_y)\frac{dz}{z}$ ?

My work so far:$$\psi_y=\frac{i}{2r}\frac{d}{dr}\int_0^{r^2}\int_0^{2\pi}(u_x+iu_y)\rho d\rho d\theta=\frac{i}{2r}2r\int_0^{2\pi}(u_x+iu_y)r^2d\theta=ir^2\int_0^{2\pi}(u_x+iu_y)d\theta$$

But

$$\frac{1}{2}\int_{|z|=r^2}(u_x+iu_y)\frac{dz}{z}=\frac{1}{2}\int_{x^2+y^2=r^4}(u_x+iu_y)\frac{dx+idy}{x+iy}\\=\frac{1}{2}\int_{x^2+y^2=r^4}(u_x+iu_y)(x-iy)\frac{dx+idy}{x^2+y^2}$$

Let $x=r^2\cos\theta, y=r^2\sin\theta$

$$\frac{1}{2}\int_{|z|=r^2}(u_x+iu_y)\frac{dz}{z}=\frac{1}{2}\int_0^{2\pi}(u_x+iu_y)\frac{(r^2\cos\theta-ir^2\sin\theta)(-r^2\sin\theta+ir^2\cos\theta)}{r^4}d\theta\\=\frac{1}{2}\int_0^{2\pi}(u_x+iu_y)id\theta$$

So how can we have $\psi_y=\frac{1}{2}\int_{|z|=r^2}(u_x+i u_y)\frac{dz}{z}$?