Solve 𝑧 squared plus two plus 𝑖
times 𝑧 plus 𝑖 equals zero.
We use the quadratic formula,
substituting the coefficients for 𝑎, 𝑏, and 𝑐. We can simplify under the
radical. Two plus 𝑖 squared is two squared,
which is four, plus two times- two times 𝑖, which is four 𝑖, plus 𝑖 squared,
which is negative one. And from this, we subtract four
times one times 𝑖, which is four 𝑖. We see that the terms involving 𝑖
cancel. And we’re left with just four minus
one, which is three under the radical.
Notice that we have a positive
discriminant here. There’s a bit more simplification
to do. Two times one on the denominator is
just two. And we can distribute the minus
sign over the parentheses. Doing this and rearranging the
terms, we get negative two plus or minus root three minus 𝑖 over two. Our solutions are therefore 𝑧
equals negative two plus root three over two minus 𝑖 over two and 𝑧 equals
negative two minus root three over two minus 𝑖 over two.
Notice that although our equation
had a positive discriminant, we don’t get two real solutions. If a quadratic equation with real
coefficients has positive discriminant, then we get two real solutions. However, our quadratic equation has
some nonreal coefficients. And so as we’ve seen, we shouldn’t
expect it to have two real roots.