Question

So if we can find a number r satisfying r2=b2−4ac, then we can solve for z in terms of a,b,c and r as

z=−b2a±r2a.

Hopefully you can recognize the usual quadratic formula here. If b2−4ac is a positive real number, we usually just replace r with b2−4ac−−−−−−−√.

However if b2−4ac is a negative real number, or in fact a general complex number---which will happen if a,b and c are complex numbers---then there is no canonical b2−4ac−−−−−−−√, but we could still find a value r (at least approximately) satisfying r2=b2−4ac, and then the formula z=−b2a±r2a gives us the solutions (as a set): {−b+r2a,−b−r2a}.

Use this analysis to solve the complex quadratic equation (1+i)z2+2z−(3+i)=0. The solution set is

Answer 1