Answer
$z_1=-1, z_2=-i$
Work Step by Step
Step 1. Identify the coefficients: $a=1, b=1+i, c=i$
Step 2. Use the quadratic formula: $z=\frac{-1-i\pm\sqrt {(1+i)^2-4i}}{2}=\frac{-1-i\pm\sqrt {1-1+2i-4i}}{2}
=\frac{-1-i\pm\sqrt {-2i}}{2}=\frac{-1-i\pm\sqrt {2}i\sqrt i}{2}$
Step 3. Evaluation $\sqrt i$: sine $i=cos\frac{\pi}{2}+i\cdot sin\frac{\pi}{2}$, we have $i^{\frac{1}{2}}=cos\frac{\pi/2+2k\pi}{2}+i\cdot sin\frac{\pi/2+2k\pi}{2}$ with $k=0,1$ which gives $\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}i$ and $-\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i$
Step 4. Use the results of Step 3 in the equation of Step 2, we can obtain all the solutions:
$z_1=\frac{1}{2}[-1-i+\sqrt 2i(\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}i)]=\frac{1}{2}[-1-i+i-1]=-1$
$z_2=\frac{1}{2}[-1-i+\sqrt 2i(-\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i)]=\frac{1}{2}[-1-i-i+1]=-i$
$z_3=\frac{1}{2}[-1-i-\sqrt 2i(\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}i)]=\frac{1}{2}[-1-i-i+1]=-i$
$z_4=\frac{1}{2}[-1-i-\sqrt 2i(-\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i)]=\frac{1}{2}[-1-i+i-1]=-1$
Step 5. In summary, the solutions are $z_1=-1, z_2=-i$
