My attempt : Define a map $f$ from $\Bbb{Z}$ to $\Bbb{Z}[i]/\langle a+ib\rangle$ by $f(n) = n+\langle a+ib\rangle$.
Then I have shown that $f$ is ring homomorphism and kernel is $\langle a^2+b^2\rangle$.
For subjectivity let p+iq $\in \mathbb{Z}[i]$. Now a+ib $\in \langle a+ib \rangle$ implies ib+$\langle a+ib \rangle$ = -a+$\langle a+ib \rangle$. Similarly i(a+ib) $\in \langle a+ib \rangle$ implies ia+$\langle a+ib \rangle$ = b+ $\langle a+ib \rangle$.
Now a and b are relatively prime implies there exist integers u, v such that au+bv=1.
Now p+iq +$\langle a+ib \rangle$ = p+iq(au+bv) +$\langle a+ib \rangle$ =p+iqau+iqbv+$\langle a+ib \rangle$ =p+bqu-aqv+$\langle a+ib \rangle$ =$f(p+bqu-aqv) $. Hence f is surjective.
Is proof of subjectivity correct?