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 $2x^2+2y^2+x-2y=0$.

A complex number $z=x+iy$ can be plotted as the point $(x,y)$, or equivalently as the position vector from the origin $O$ to that point, in the Argand plane (named for Jean Argand) — the $x$-axis is the real axis and the $y$-axis is the imaginary axis. This geometric picture turns algebraic facts about $z$ into statements about points, lines and circles.

Distance and circles. Since $|z_1-z_2|$ is the distance between the points $z_1$ and $z_2$, the equation

is exactly the set of points at distance $r$ from the fixed point $z_0$ — i.e. the complex form of a circle with centre $z_0$ and radius $r$. Correspondingly:

• $|z-z_0|<r$ describes the interior of that circle;
• $|z-z_0|>r$ describes the exterior. An equation like $|\alpha z-\beta|=\gamma$ ($\alpha\ne0$) is first rewritten as $|z-(\beta/\alpha)|=\gamma/|\alpha|$ to read off centre $\beta/\alpha$ and radius $\gamma/|\alpha|$.

Loci from conditions on $z,\overline z$. Many geometric conditions translate directly:

• $|z-a|=|z-b|$ (equidistant from two fixed points) is the perpendicular bisector of the segment joining $a,b$ — e.g. $|z+2|=|z-2|$ gives the imaginary axis, since $\pm2$ are symmetric about it.
• Conditions like $\operatorname{Re}(iz)=k$, $\operatorname{Im}[(a+ib)z+c]=0$, or $\overline z=z^{-1}$ each reduce, after substituting $z=x+iy$ and separating real/imaginary parts, to an ordinary Cartesian equation in $x,y$ — a line, a circle, or (as with $\overline z=z^{-1}\iff z\overline z=1\iff x^2+y^2=1$) the unit circle.
• An argument condition, e.g. $\arg\!\left(\dfrac{z-i}{z+2}\right)=\dfrac\pi4$, fixes the angle a chord subtends and typically traces an arc of a circle; converting to Cartesian form again proceeds by writing $z=x+iy$, isolating the real/imaginary parts of the quotient, and using $\tan(\text{angle})$.