Đặt $z=r(cos\frac{\pi}{4}+isin\frac{\pi}{4})=\frac{r}{\sqrt2}+i.\frac{r}{\sqrt2}$$z+2+3i=(\frac{r}{\sqrt2}+2)+i(\frac{r}{\sqrt2}+3)$
$(|z+2+3i|)^2=(\frac{r}{\sqrt2}+2)^2+(\frac{r}{\sqrt2}+3)^2=25$
Đặt $\frac{r}{\sqrt2}=a>0$
$\Rightarrow (a+2)^2+(a+3)^2=25$
$\Rightarrow 2a^2+10a-12=0$
$\Rightarrow a=1$
$\Rightarrow r=\sqrt2$