Q.Show that the following equations represent a circle, and, find its centre and radius.

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. …

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 $1,2,-7$). …

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. …

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 $e^{i\pi/2}=\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}=i$. …

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: $4x^2-4x+1+4y^2 = 4x^2+4y^2$. …

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 $y<0$. …

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): $\dfrac{z_1+z_3}{2}=\dfrac{z_2+z_4}{2}\ \Longrightarrow\ z_1+z_3=z_2+z_4$. …

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. …