## Calculus: Early Transcendentals 8th Edition

$-\frac{1}{4}\pm\frac{\sqrt{3}}{4}i$
$z^{2}+\displaystyle \frac{1}{2}z+\frac{1}{4}=0$ First we multiply through by 4: $4z^{2}+2z+1=0$ Next, we solve using the quadratic formula (a=4, b=2, c=1): $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $z=\displaystyle \frac{-2\pm\sqrt{2^{2}-4(4)(1)}}{2(4)}$ $=\frac{-2\pm\sqrt{4-16}}{8}$ $=\frac{-2\pm\sqrt{-12}}{8}$ $=\frac{-2\pm\sqrt{-1*3*4}}{8}$ We use the fact that $\sqrt{-1}=i$: $=\frac{-2\pm 2\sqrt{3}i}{8}$ $=-\frac{1}{4}\pm\frac{\sqrt{3}}{4}i$