Q.If the real part of $\dfrac{\bar{z}+2}{\bar{z}-1}$ is 4, then show that the locus of the point representing $z$ in the complex plane is a circle.

The key idea is to convert the given complex number condition into a Cartesian equation using the standard form of a circle.

1. Let $z = x + iy$, where $x$ and $y$ are real numbers. Then $\bar{z} = x - iy$.
2. Substitute $\bar{z}$ into the given expression:

3. To find the real part, multiply the numerator and denominator by the conjugate of the denominator:

The real part is $\text{Re}\left(\frac{\bar{z}+2}{\bar{z}-1}\right) = \frac{x^2+y^2+x-2}{(x-1)^2+y^2}$.

4. Given that the real part is 4:

Assuming $(x-1)^2+y^2 \neq 0$ (i.e., $z \neq 1$), we multiply both sides by the denominator:

Rearranging the terms to one side:

Dividing the entire equation by 3: …

The problem asks us to determine the geometric path (locus) traced by a point $z$ in the complex plane, given a specific condition on an expression involving $z$. The most direct way to approach such problems is to express the complex number $z$ in its Cartesian form, $z = x + iy$, and then translate the given complex condition into an equation involving $x$ and $y$. If this resulting equation matches the standard form of a circle, $(x-h)^2 + (y-k)^2 = r^2$, then we have successfully shown that the locus is a circle.

Let's proceed step-by-step:

1. Represent $z$ in Cartesian form.

   Let $z = x + iy$, where $x$ and $y$ are real numbers representing the coordinates of the point in the complex plane.

   The conjugate of $z$, denoted $\bar{z}$, is then $x - iy$.

2. Substitute into the given expression.

   The given expression is $\dfrac{\bar{z}+2}{\bar{z}-1}$.

   Substitute $\bar{z} = x - iy$ into the expression:

3. Simplify the complex fraction. To find the real part of this complex fraction, we need to eliminate the complex term from the denominator. We do this by multiplying both the numerator and the denominator by the conjugate of the denominator. The denominator is $(x-1) - iy$, so its conjugate is $(x-1) + iy$.

Now, we expand the numerator and the denominator:
*   **Numerator:**
    $((x+2) - iy)((x-1) + iy)$
    $= (x+2)(x-1) + (x+2)(iy) - iy(x-1) - (iy)(iy)$
    $= (x^2 + x - 2) + i(xy + 2y) - i(xy - y) - i^2y^2$
    $= (x^2 + x - 2) + i(xy + 2y - xy + y) + y^2 \quad (\text{since } i^2 = -1)$
    $= (x^2 + x - 2 + y^2) + i(3y)$

*   **Denominator:**
    $((x-1) - iy)((x-1) + iy)$
    This is in the form $(A-B)(A+B) = A^2 - B^2$. Here, $A = (x-1)$ and $B = iy$.
    So, $(x-1)^2 - (iy)^2 = (x-1)^2 - i^2y^2 = (x-1)^2 - (-1)y^2 = (x-1)^2 + y^2$.

Combining these, the simplified expression is:

4. Identify the real part. We can separate the real and imaginary parts of the simplified expression:

5. Set the real part equal to 4. The problem states that the real part of the expression is 4:

> [!WARNING]
> The denominator $(x-1)^2 + y^2$ cannot be zero. If it were, it would imply $x-1=0$ and $y=0$, meaning $x=1, y=0$. This corresponds to $z=1$. If $z=1$, the original expression $\frac{\bar{z}+2}{\bar{z}-1}$ becomes $\frac{1+2}{1-1} = \frac{3}{0}$, which is undefined. Therefore, the point $z=1$ (or $(1,0)$ in Cartesian coordinates) is excluded from the locus.

6. Simplify the equation to the standard form of a circle.

Multiply both sides by the denominator: …