solving complicated complex numbers

Question

Find values for $a$ and $b$ so that $z=a+bi$ satisfies $\displaystyle \frac{z+i}{z+2}=i$. Below are my workings:

so far i simplify $\displaystyle \frac{z+i}{z+2}=i$ to $z=zi+i$

which $a=i$, $b=z$

Answer 1

$$\dfrac{z+i}{z+2}=i\\ \implies a=-b\mbox{ and }\\ \mbox{}b+1=a+2\mbox{ (by taking the imaginary and real parts of the resulting equation.)}$$ Solving gives $$\boxed{b=0.5,a=-0.5}$$

Answer 2

Hint: You have $\frac {z+i}{z+2}=i$ Multiplying by $z+2$ gives $z+i=iz+2i$ Now substitute in $z=a+bi$, remembering that $a$ and $b$ are real. The real and imaginary parts give you two equations in the two unknowns $a,b$

Answer 3

$z=zi+i$ is equivalent to $z-zi=i$. $z(1-i)=i$ then $z=\frac{i}{1-i}$ so $a = Re(z)$ and $b=Im(z).$