We have

$(a^2+b^2)(c^2+d^2) = a^2 c^2+a^2d^2+b^2c^2+b^2d^2$

$(ac+bd)^2=a^2c^2+2abcd+b^2d^2=0$

Subtracting the two gives us $(ad-bc)^2=\pm 1$. WLOG $ad-bc=1$. Then

\[ (a-d)^2+(b+c)^2=a^2-2ad+d^2+b^2+2bc+c^2=(a^2+b^2)+(c^2+d^2)-2(ad-bc)=0. \]