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 $a=i$, $b=z$
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$a$ and $b$ are supposed to be real numbers, not $i$ nor $z$ – Ross Millikan Mar 11 '14 at 23:37
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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 $a,b$
Ross Millikan
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No, substitute $a+bi$ for $z$ in the previous equation. It becomes $a+bi+i=i(a+bi)+2i$ – Ross Millikan Mar 11 '14 at 23:46
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@harry Algebra still works the same way for complex numbers as it does for reals. The added "feature" is that inserting $ \ z = a + bi \ $ into the algebraic equation $ \ z + i \ = \ iz + 2i \ $ produces real and imaginary parts on both sides of the equation, which must be "matched up" separately. (We do this with real-number algebra all the time: we just don't think about it as "making parts agree" because the "imaginary part" is always zero...) – colormegone Mar 12 '14 at 01:08
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$$\dfrac{z+i}{z+2}=i\\ \implies a=-b\mbox{ and }\\ \mbox{}b+1=a+2\mbox{ (by taking the imaginary and real parts of the resulting equation.)}$$ Solving gives $$\boxed{b=0.5,a=-0.5}$$
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@harry As I've said, you can use the real and imaginary parts of the equation you get when you write $z+i=iz+2i$. – Mar 11 '14 at 23:48
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So, let me multiply it out: $a+bi+i=ia-b+2i$. Take the real and imaginary part - and you get the desired equations. – Mar 11 '14 at 23:50
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As I've said, solve for $b,a$ by taking the real and imaginary parts of $a+bi+i=ia-b+2i$. – Mar 11 '14 at 23:55
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$z=zi+i$ is equivalent to $z-zi=i$. $z(1-i)=i$ then $z=\frac{i}{1-i}$ so $a = Re(z)$ and $b=Im(z).$
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