Show that Re((1)/(z)-(2)/(3))=0 represents the equation of a…
Show that $\text{Re}\left(\dfrac{1}{z}-\dfrac{2}{3}\right)=0$ represents the equation of a circle.
Explanation of Method
Rewrite the condition as $\text{Re}\left(\frac1z\right)=\frac23$, substitute $z=x+iy$, and rationalize $\frac1z$. The resulting equation in $x,y$ is a first-degree equation in $x^2+y^2$, $x$, $y$ — which, after completing the square, is exactly the equation of a circle.
Step 1 · Isolate $\text{Re}(1/z)$
$$ \text{Re}\left(\frac1z\right) - \frac23 = 0 \quad\Longrightarrow\quad \text{Re}\left(\frac1z\right) = \frac23 $$
Step 2 · Rationalize $\dfrac1z$
With $z=x+iy$:
$$ \frac1z = \frac1{x+iy}\cdot\frac{x-iy}{x-iy} = \frac{x-iy}{x^2+y^2} \quad\Longrightarrow\quad \text{Re}\left(\frac1z\right) = \frac{x}{x^2+y^2} $$
Step 3 · Apply the condition
$$ \frac{x}{x^2+y^2} = \frac23 $$
Step 4 · Clear denominators
$$ 3x = 2(x^2+y^2) \quad\Longrightarrow\quad 2x^2-3x+2y^2=0 \quad\Longrightarrow\quad x^2-\frac32 x+y^2=0 $$
Step 5 · Complete the square
$$ \left(x-\frac34\right)^2-\frac9{16}+y^2=0 \quad\Longrightarrow\quad \boxed{\left(x-\frac34\right)^2+y^2=\left(\frac34\right)^2} $$
Circle of radius $3/4$ centered at $(3/4,0)$, tangent to the imaginary axis at the origin (which is excluded, since $1/z$ is undefined there).
Step 6 · Properties of the circle
| Property | Value |
|---|---|
| Center | $\left(\frac34,0\right)$ |
| Radius | $\frac34$ |
| Tangent to | the imaginary axis, at the origin |
| Rightmost point | $\left(\frac32,0\right)$ |
| Excluded point | the origin ($1/z$ undefined at $z=0$) |
Step 7 · Verification
At $z=\frac32$: $\ \text{Re}\left(\frac{1}{3/2}\right)=\text{Re}\left(\frac23\right)=\frac23$, so $\text{Re}\left(\frac1z-\frac23\right)=0$. Check on the circle: $\left(\frac32-\frac34\right)^2+0^2=\left(\frac34\right)^2$ ✓
At $z=i$: $\ \frac1z=-i$, so $\text{Re}(-i)=0\neq\frac23$, and indeed $(0-\frac34)^2+1^2=\frac{25}{16}\neq\frac9{16}$ — $z=i$ is correctly excluded. ✓
Step 8 · Why the circle is tangent to the imaginary axis at the origin
The leftmost point of the circle has $x=\frac34-\frac34=0$, so the circle touches the $y$-axis only at the origin — the very point excluded from the domain of $\frac1z$. This is a general feature of the transformation $w=\frac1z$: it maps the vertical line $\text{Re}(w)=k$ (which misses the origin, for $k\neq0$) to a circle through the origin.
Result
A circle centered at $\left(\frac34,0\right)$ with radius $\frac34$, tangent to the imaginary axis at the (excluded) origin.