Complex Numbers: Im$(\frac{12}{z-7})=1$

Question

Sketch and describe the set of complex numbers satisfying $$Im(\frac{12}{z-7})=1$$

where $z=x+iy$

The answer should be in circle form.

Here is what I have so far:

$$Im(12)=z-7$$ $$Im(12)=x+iy-7$$

At this point I am now stuck and not sure what to do, or I have gone about this the wrong way?

Answer 1

The imaginary part of $a / b$ is not the imaginary part of $a$ over $b$, which is what you've used in your first step. So, yes, you have gone the wrong way.

I suggest that you change $$ \frac{12}{z-7} $$ by multiplying both top and bottom by $\bar{z} - 7$; the denominator will then be real, and you can bring that outside the "Im". See where that gets you, OK? (Also: go ahead and write $z = x + iy$, to make it easier to tell what's real and what's imaginary.) If you need more help after this, just ask.

Answer 2

You need to start with $z=x+iy$ and then compute the imaginary part of $\frac{12}{z-7}$.

\begin{align*} \frac{12}{z-7} & = \frac{12}{(x-7)+iy}\\ & = 12\frac{[(x-7)-iy]}{(x-7)^2+y^2} \end{align*} Thus $$\text{Im}\frac{12}{z-7}=\frac{-12y}{(x-7)^2+y^2}.$$ Now proceed from here.

Answer 3

You can try this way: let $z=x+iy$, so $\displaystyle \Im \frac{12}{z-7}=\Im \frac{12}{x+iy-7}=\Im \frac{12(x-7-iy)}{(x-7)^2+y^2}=\frac{-12y}{(x-7)^2+y^2}=1$, so:

$$(x-7)^2+y^2=-12y$$

It's a circle equation.

Answer 4

Rewrite as $\dfrac{12}{(x - 7) + yi}$

Multiply top and bottom by complex conjugate $(x-7) - yi$

$\dfrac{12x - 84 - 12yi}{(x-7)^2 + y^2}$

We want the imaginary part of this to be equal to $1$.

$\dfrac{-12y}{(x-7)^2 + y^2} = 1$

$y^2 + 12y + (x-7)^2 = 0$

Complete the square:

$(y + 6)^2 + (x-7)^2 = 36$

This is the equation of a circle centered at $(7, -6)$ with radius $6$.