Put \(z=x+iy\) and multiply the fraction by the conjugate of its denominator. Its imaginary numerator is \[(y+5)(x-5)-xy=5x-5y-25.\] The denominator is real and non-zero, since \(|z|=\sqrt{17}\) excludes \(z=5\). Thus \(y=x-5\). The modulus gives \[x^2+(x-5)^2=17\Rightarrow x^2-5x+4=0.\] Hence \(x=4\) or 1, with \(y=-1\) or \(-4\). Therefore \[z=4-i\quad\text{or}\quad z=1-4i.\]