I am unsure of how to solve the following question. I provided my attempt however I am doubtful that it is correct.
Determine the image of the lines $Re(z) = a$ and $ℑ(z) = b$ under the transformation $f(z) = \frac 1 z$
My attempt: $z=x+iy$, but if $Re(z)=a$ therefore $z=a+iy$
since $f(z)=\frac 1 z$, we obtain $\frac 1 {a+iy}$
Multiplying the fraction by its complex conjugate we obtain $\frac {a-iy} {a^2+y^2}$
$f(z)= u(a,y)+iv(a,y)$
where $u(a,y)= \frac a {a^2+y^2}$ and $v(a,y)= \frac {-y} {a^2+y^2}$
Unsure of: $u=\frac {-a}{y}.v$ so $y=\frac{-a.v}{u}$
Thus, $u^2+v^2-\frac u a =0$
So it forms a circle of radius $\sqrt \frac u a$ centered at $(\frac u a , 0)$
Would i use the same method for $ℑ(z) = b$ or should I have done them as a combination to begin with?
Any help would be great. Thanks in advance!
Note that $$u^2+v^2-\frac{u}{a} = u^2 - 2\cdot u \cdot\frac{1}{2a} + \frac{1}{4a^2} + v^2 - \frac{1}{4a^2} = \\= \left(u - \frac{1}{2a}\right)^2 + v^2 - \frac{1}{4a^2},$$ thus the equation of the image of the line $\operatorname{Re}(z)=a$ is $$\left(u - \frac{1}{2a}\right)^2 + v^2 = \frac{1}{4a^2}.$$