Transformations from the z-plane to the w-planeEdexcel International A Level Further Maths: Revision notes

A transformation $T$ sends each point $z=x+iy$ in the $z$-plane to a point $w=u+iv$ in the $w$-plane, where $w=f(z)$. Points, lines, circles and regions in the $z$-plane have images in the $w$-plane. There are two standard methods for the image of a locus:

• Write $w$ in terms of $x$ and $y$, separate real and imaginary parts to get $u$ and $v$, then eliminate $x$ and $y$.
• Rearrange to give $z$ in terms of $w$, substitute into the equation of the locus, and simplify to a relation between $u$ and $v$. This is usually easier when $w$ is a quotient. Moduli and arguments help: $|w|=|f(z)|$ and $\arg w=\arg f(z)$. Recall that $|z-a|=r$ is a circle (centre $a$, radius $r$) and $|z-a|=|z-b|$ is the perpendicular bisector of the points $a$ and $b$.

For $w=az+b$ with $a,b$ complex and $a\neq0$:

• $|a|$ is the scale factor of an enlargement about the origin;
• $\arg a$ is the angle of rotation about the origin (anticlockwise if positive);
• $b$ is a translation. Lines map to lines and circles to circles. A circle $|z-z_0|=r$ maps to the circle $|w-(az_0+b)|=|a|r$: the centre moves to $az_0+b$ and the radius is multiplied by $|a|$. Example: $w=(1+i)z-2$ has $|1+i|=\sqrt2$ and $\arg(1+i)=\frac{\pi}{4}$, so it is an enlargement by $\sqrt2$, a rotation of $\frac{\pi}{4}$ anticlockwise, then a translation of $-2$. The circle $|z-1|=2$ maps to the circle with centre $(1+i)(1)-2=-1+i$ and radius $2\sqrt2$.

Write $z=re^{i\theta}$. Then $w=z^2=r^2e^{2i\theta}$, so $|w|=|z|^2$ and $\arg w=2\arg z$.

• The circle $|z|=r$ maps to the circle $|w|=r^2$.
• The half-line $\arg z=\alpha$ maps to the half-line $\arg w=2\alpha$.
• The points $z$ and $-z$ have the same image, so the map is two-to-one. For other lines use $z=x+iy$: $w=x^2-y^2+2ixy$, so $u=x^2-y^2$ and $v=2xy$. Example: the line $x=3$ has $u=9-y^2$ and $v=6y$. Eliminating $y$ gives the parabola $v^2=36(9-u)$. Example: the line $y=x$ has $u=0$ and $v=2x^2\ge0$, so it maps to the non-negative imaginary axis.

A transformation $w=\frac{az+b}{cz+d}$ with $a,b,c,d$ complex (and $ad-bc\neq0$) is undefined at the pole $z=-\frac dc$. Rearrange to find $z$ in terms of $w$: $z=\frac{dw-b}{a-cw}.$ Substitute into the locus and simplify. In general:

• a line or circle through the pole maps to a line;
• a circle not through the pole maps to a circle;
• a line not through the pole maps to a circle (through the point $w=\frac ac$). Example: $w=\frac1z$ and the line $\mathrm{Im}\,z=1$. Then $z=x+i$ and $w=\frac{x-i}{x^2+1}$, so $u=\frac{x}{x^2+1}$ and $v=\frac{-1}{x^2+1}$. Hence $u^2+v^2=\frac{1}{x^2+1}=-v$, which is the circle $u^2+\left(v+\frac12\right)^2=\frac14$ through the origin. Example: $w=\frac1z$ and the circle $|z|=2$ gives $|w|=\frac12$.

To find the image of a region (an inequality), find the image of its boundary, then decide which side. Either substitute $z$ in terms of $w$ directly into the inequality, or test one point inside the original region and see where it lands. Example: under $w=\frac{1}{z-2}$ the interior $|z-1|<1$ gives $|w+1|<|w|$, that is $u<-\frac12$. The test point $z=1$ maps to $w=-1$, which satisfies $u<-\frac12$. For a half-plane such as $\mathrm{Im}\,z>0$, write $\mathrm{Im}\,z$ in terms of $u$ and $v$ before applying the inequality. Take care when multiplying an inequality by an expression such as $u^2+v^2$: it is positive, so the direction is kept. Summary of the method: (1) name the locus, (2) express $z$ in terms of $w$ or $u,v$ in terms of $x,y$, (3) substitute, (4) simplify to $u$ and $v$, (5) identify the curve and state any excluded point.