Q.If $z=x+iy$ is a complex number such that $\operatorname{Im}\left(\dfrac{2z+1}{iz+1}\right)=0$, show that the locus of $z$ is $2x^2+2y^2+x-2y=0$.

If $-x-iy$ lies in the first quadrant then $-ix+y$ lies in the second quadrant.

1. A complex number lies in the first quadrant when both its real and imaginary parts are positive.
2. $-x-iy$ has real part $-x$ and imaginary part $-y$. First quadrant $\Rightarrow -x>0$ and $-y>0$, i.e. $x<0$ and $y<0$.
3. Now consider $-ix+y = y - ix$, whose real part is $y$ and imaginary part is $-x$.
4. Since $y<0$, the real part of $y-ix$ is negative.
5. Since $x<0$, we have $-x>0$, so the imaginary part of $y-ix$ is positive.
6. A complex number with negative real part and positive imaginary part lies in the second quadrant.

The three points are real-number multiples of the same complex number $z_1z_2z_3$, so they are collinear.

1. Compute $z_1z_2=(1+2i)(1-3i)=1-3i+2i-6i^2=1-i+6=7-i$.
2. Compute $z_1z_2z_3=(7-i)(2+4i)=14+28i-2i-4i^2=14+26i+4=18+26i$.
3. Let $w=z_1z_2z_3=18+26i$. The three given points are $w$, $2w$, and $-7w$.
4. Each of these is a real scalar multiple of the same complex number $w$ (multiples $1,2,-7$).
5. On the Argand plane, all real scalar multiples of a fixed complex number $w\neq0$ lie on the single line through the origin passing through the point $w$.
6. Hence the points representing $w$, $2w$, $-7w$ all lie on this one line, i.e. they are collinear.

The equation states that $z$ is equidistant from the fixed points $z_1$ and $z_2$, which is exactly the geometric definition of the perpendicular bisector of the segment $z_1z_2$.

1. Interpret $|z-z_1|$ as the distance from the point $z$ to the fixed point $z_1$, and $|z-z_2|$ as the distance from $z$ to the fixed point $z_2$.
2. The equation $|z-z_1|=|z-z_2|$ says these two distances are always equal.
3. The locus of points equidistant from two fixed points $z_1,z_2$ is, by the standard geometric definition, the perpendicular bisector of the segment joining them.
4. It is a straight line, but not through the origin in general (rules out (a)), and it is not a circle (rules out (b), (d)) unless $z_1=z_2$ trivially.

Rotating the point $iz$ counter-clockwise through $\pi/2$ (multiplying by $i$) gives $i(iz)=-z$.

1. Rotating a complex number $w$ counter-clockwise about the origin through an angle $\theta$ corresponds to multiplying it by $e^{i\theta}$.
2. Here $\theta=\dfrac{\pi}{2}$, so $e^{i\pi/2}=\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}=i$.
3. The starting point is $w=iz$.
4. After rotation, the new point is $i\cdot w = i\cdot(iz) = i^2 z$.
5. Since $i^2=-1$, this equals $-z$.

Substitute $z=x+iy$, square both sides of $|2z-1|=2|z|$, and simplify — the $y^2$ terms cancel, leaving a vertical line $x=1/4$.

1. Let $z=x+iy$, so $2z-1 = (2x-1)+2iy$.
2. $|2z-1|=2|z|$ means $\sqrt{(2x-1)^2+(2y)^2} = 2\sqrt{x^2+y^2}$.
3. Square both sides: $(2x-1)^2+4y^2 = 4(x^2+y^2) = 4x^2+4y^2$.
4. Expand the left side: $4x^2-4x+1+4y^2 = 4x^2+4y^2$.
5. The $4x^2$ and $4y^2$ terms cancel from both sides: $-4x+1=0$.
6. Solve: $x=\dfrac14$.
7. This is a vertical straight line, matching option (a).

If $-\bar z$ lies in the third quadrant, then $z$ lies in the fourth quadrant.

1. Let $z=x+iy$. Then $\bar z=x-iy$ and $-\bar z=-x+iy$.
2. $-\bar z$ lying in the third quadrant means both its real and imaginary parts are negative: $-x<0$ and $y<0$.
3. $-x<0\Rightarrow x>0$; and $y<0$.
4. So $z=x+iy$ has $x>0,\ y<0$, which is exactly the fourth quadrant.
5. Options (a), (b), (c) correspond to other sign combinations that do not match $x>0,\,y<0$.

The parallelogram condition is exactly the diagonals-bisect-each-other condition on the two diagonal vertex pairs.

1. For vertices $z_1,z_2,z_3,z_4$ taken in order, the sides are $z_1z_2$, $z_2z_3$, $z_3z_4$, $z_4z_1$, and the diagonals are $z_1z_3$ and $z_2z_4$.
2. A quadrilateral is a parallelogram if and only if its diagonals bisect each other (a standard geometric characterisation, equally valid in the complex plane, where midpoints are just averages).
3. Midpoint of diagonal $z_1z_3$ is $\dfrac{z_1+z_3}{2}$; midpoint of diagonal $z_2z_4$ is $\dfrac{z_2+z_4}{2}$.
4. Setting these equal (bisection): $\dfrac{z_1+z_3}{2}=\dfrac{z_2+z_4}{2}\ \Longrightarrow\ z_1+z_3=z_2+z_4$.
5. Checking the distractors: (a) $z_1+z_4=z_2+z_3$ and (c) $z_1+z_2=z_3+z_4$ pair up adjacent (side) vertices, not the diagonal pairs, so they don't correspond to the diagonal-bisection condition; (d) is merely a side-equality/parallel-sides condition on one pair, not the full iff condition.

Equidistance from two fixed points is the geometric definition of the perpendicular bisector, so that is the locus.

1. Let $z=x+iy$, $z_1,z_2$ be fixed complex numbers (points in the plane).
2. $|z-z_1|$ is the distance from the variable point $z$ to the fixed point $z_1$; likewise $|z-z_2|$ is the distance to $z_2$.
3. The condition $|z-z_1|=|z-z_2|$ says the point $z$ is always equidistant from $z_1$ and $z_2$.
4. In plane geometry, the set of all points equidistant from two fixed points is, by definition, the perpendicular bisector of the segment joining those two points.
5. This is not centred at the origin or at $z_1$ (ruling out (a), (b)), and it need not pass through the origin at all (ruling out (c)) — it is simply the perpendicular bisector of $z_1z_2$, which may lie anywhere in the plane.