Video Transcript
Solve 𝑧 squared plus two minus two
𝑖 𝑧 minus seven plus 26𝑖 equals zero.
We use the quadratic formula. We substitute the values of the
coefficients for 𝑎, 𝑏, and 𝑐. We simplify under the radical sign,
noticing there’s some cancellation. And negative eight 𝑖 plus 28 plus
104𝑖 is 28 plus 96𝑖. We notice that, under the radical,
we have two multiples of four. So we can take out that multiple of
four in each case and bring out a two in front of the radical sign. This allows us to cancel the two in
the denominator. So we get negative one plus 𝑖 plus
or minus the square root of seven plus 24𝑖.
We use de Moivre’s theorem to find
the square roots. But first, we need to write seven
plus 24𝑖 in polar form. Its modulus is 25 and its argument
is arctan 24 over seven. And de Moivre’s theorem for roots
tells us how to find the square roots.
The modulus of the square root is
the square root of the modulus. So that’s the square root of 25,
which is five. We have to halve the argument. The argument of the square root is
arctan 24 over seven divided by two. Putting these into a calculator, we
find that cos of this value is 0.8 and sin of this value is 0.6. And so upon multiplying by the
modulus five, we add or subtract four plus three 𝑖. Therefore, our two roots are 𝑧
equals three plus four 𝑖 and 𝑧 equals negative five minus two 𝑖.
Here, as the discriminant was
complex, we had to use de Moivre’s theorem to find its square roots. Actually, we only found one of its
square roots using de Moivre’s theorem. But we knew that the other must be
its opposite. And we have this plus or minus sign
here.
