Solve the equation: \$ z^2 + 6z + 13 = 0\$, where z is a complex number.

To solve the quadratic equation \$z^2 + 6z + 13 = 0 \$, where z is a complex number, we can use the quadratic formula:

\$ z = ((-b) pm \sqrt(b^2 - 4ac))/(2a) \$

Where a = 1, b = 6, and c = 13 in our equation.

Let’s calculate the values of z

\$ z = ((-6) pm \sqrt(6^2 - 4(1)(13)))/(2(1)) \$

\$ z = ((-6) pm \sqrt(- 16))/(2) \$

Since the discriminant (\$b^2−4ac\$) is negative, we get a complex square root of −16, which is 4i. Therefore:

\$ z = (- 6 pm 4i)/(2) \$

Simplifying each term:

\$ z = −3 pm 2i \$

z = - 3 + 2i and z = - 3 - 2i

This option doesn’t match our solutions because one of the roots should have a negative real part (-3), but the second root has a positive real part (3)

z = -3 + 2i and z = 3 + 2i

This option doesn’t match our solutions because the first root has a positive real part (3), but one of the roots should have a negative real part (-3)

z = 3 + 2i and z = -3 - 2i

This option matches our solutions. Both roots have the correct imaginary part (2i and -2i), and one has the correct negative real part (-3), while the other has the same negative real part (-3)

z = -3 + 2i and z = -3 - 2i

This option doesn’t match our solutions because both roots have positive real parts (3), but one of the roots should have a negative real part (-3)

z = 3 + 2i and z = 3 - 2i

Option

C

Can you summarize what you’ve understood in the above steps?