any ideas?

Jjesus wrote: any ideas? Consider $u:=a+bi, v:=c-di$. The condition becomes $uv=90121(u+v)$ or \[ (u-90121)(v-90121)=90121^2=(300+11i)^2(300-11i)^2. \]Note that $300\pm11i$ are prime in $\mathbb{Z}[i]$. The rest is a simple case distinction.

We have $ac+pd = p(a+c)$, yielding $(a-p)(c-p) = p^2-bd$. Likewise, $bc-ad=p(b-d)$ yields $b(c-p) = d(a-p)$. Suppose first $a=p$. Then $c=p$, yielding $bd=p^2$. It is not hard to see both $(b,d)=(1,p^2)$ and $(p^2,1)$ work, so suppose $a,c\ne p$. Let $z={\rm gcd}(b,d)$ and $b=zb_1,d=zd_1$. We obtain $b_1(c-p)=d_1(a-p)$. In particular, $a-p$ and $c-p$ has the same sign and non-zero, so $p^2-bd>0$. As $(b_1,d_1)=1$, we have $b_1\mid a-p$ and $d_1\mid c-p$. Returning to $(a-p)(c-p)=p^2-bd$, we find $b_1d_1\mid (a-p)(c-p)$ due to $(b_1,d_1)=1$, so $b_1d_1\mid p^2$. Since $bd = z^2b_1d^2<p^2$, the possibilities are $b_1d_1=1$ or $b_1d_1=p$. If $b_1d_1=1$ then $a=c$; this yields $c=a=p$, a case that's already inspected. If $b_1d_1=p$, then we can assume $b_1=p$ and $d_1=1$ (the other case is analogous). We have $p(c-p)=(a-p)$ so $p\mid a-p$. Set $a=kp$. Then $c=p+k-1$, $b=pz$ and $d=z$. Furthermore, upon plugging in we find $(k-1)^2 +z^2 = p$. This yields $\{k-1,z\} = \{300,11\}$, due to the uniqueness of prime representation as sum of two squares. The rest is boring casework, and omitted.