FP2 4.2 - Transformations from z-plane to w-plane | Edexcel A-Level Further Mathematics

Find complete images of points, loci and regions under complex transformations (FP2 4.2). You will square in Cartesian or polar form, invert fractional transformations and preserve poles, missing points and boundary restrictions.

Squaring can bend a straight line

How can a straight line become a parabola, while another simple transformation sends a line onto a circle with one point missing? A complex formula changes whole sets of points. Finding an equation for the image is only part of finding its complete image.

You already know

Write $z=x+\mathrm{i}y$ in the source plane and $w=u+\mathrm{i}v$ in the image plane. Complex multiplication can be calculated by expansion or polar form. A locus and its domain exclusions must be kept together; argument conditions always exclude a zero displacement.

Expanding $z^2$ gives two ordinary real equations.

Squaring in Cartesian coordinates

For $w=z^2$ ,

u=x^2-y^2,\qquad v=2xy

The origin maps to the origin. For $w\ne0$ , the two square roots $z$ and $-z$ have the same image; the origin is its own single preimage.

Not in the formula booklet: learn it.

For a power map, polar form explains the change in modulus and direction. The underlying theorem is supplied.

De Moivre's theorem

For $z=r(\cos\theta+\mathrm{i}\sin\theta)$ and positive integer $n$ ,

z^n=r^n(\cos n\theta+\mathrm{i}\sin n\theta)

In the formula booklet.

Thus $w=z^n$ sends modulus $r$ to $r^n$ . For $z\ne0$ , its angle becomes $n\theta$ , then is reduced to $(-\pi,\pi]$ if a principal argument is requested. These image interpretations are not separately printed. They do not describe a single fixed enlargement or rotation of the plane.

Worked example 1: the image of a vertical line

Question. Under $w=z^2$ , find the complete image of the line $\Re z=1$ .

Plan. Set $x=1$ , then eliminate $y$ between the equations for $u,v$ .

\begin{aligned} u&=1-y^2,\qquad v=2y &&\text{substitute the line condition}\\ y&=\frac v2 &&\text{solve for the source parameter}\\ u&=1-\frac{v^2}{4} &&\text{eliminate that parameter} \end{aligned}

Why this step? Every real $v$ gives a real source parameter $y=v/2$ , establishing the whole parabola rather than just a necessary equation.

Answer. The complete parabola $u=1-v^2/4$ , for all real $v$ .

Check. $z=1+\mathrm{i}$ maps to $2\mathrm{i}$ , giving $u=0,v=2$ , which satisfies the parabola.

Change only the source line from $\Re z=1$ to $\Re z=2$ .

Worked example 2: a shifted source line

Question. Find the image of $\Re z=2$ under $w=z^2$ .

Plan. Substitute $x=2$ , then eliminate the new relationship between $v$ and $y$ .

\begin{aligned} u&=4-y^2,\qquad v=4y &&\text{new fixed real part}\\ y&=\frac v4 &&\text{recover the source parameter}\\ u&=4-\frac{v^2}{16} &&\text{complete elimination} \end{aligned}

Why this step? The coefficient of $v^2$ changes as well as the vertex; this is not merely a translation of the earlier image.

Answer. The complete parabola $u=4-v^2/16$ , for all real $v$ .

Check. $z=2+\mathrm{i}$ maps to $3+4\mathrm{i}$ , and $3=4-16/16$ .

A line through the origin is a boundary case worth checking directly: $x=0$ gives $v=0,\ u=-y^2\leq0$ , including the origin if the original line includes $z=0$ .

Watch out

The statement “ $w=z^2$ enlarges every distance by factor $2$ ” is wrong: modulus $3$ becomes $9$ , rather than $6$ . Moduli are squared and arguments doubled; the changes depend on the original point.

Change the fixed real part once more.

Map radii and angles, then choose principal directions

For an annular sector, power maps separate neatly into a radial interval and an angular interval. Keep the angle temporarily unwrapped so that crossing the principal cut does not discard an edge.

Squaring an annular sector

For positive radii $r_1\leq|z|\leq r_2$ and an allowed source angle $\theta$ , squaring gives

r_1^2\leq|w|\leq r_2^2,\qquad w=|z|^2(\cos2\theta+\mathrm{i}\sin2\theta)

Map the entire angle interval before expressing the answer in principal arguments. Boundary inclusion follows from which source points are actually allowed.

Not in the formula booklet: learn it.

A full circle $|z|=r$ maps to the full circle $|w|=r^2$ . A restricted source arc may cover it once, twice or only partly; check the actual angle interval.

Worked example 3: an upper half-annulus

Question. Find the image under $w=z^2$ of

1\leq|z|\leq2,\qquad0\leq\arg z\leq\frac\pi2

Plan. Square the radius bounds and double every allowed angle.

\begin{aligned} 1\leq|w|&\leq4 &&\text{square the radial bounds}\\ 0\leq2\theta&\leq\pi &&\text{double the angle interval}\\ \Im w&\geq0 &&\text{upper-half directions} \end{aligned}

Why this step? Any image radius between $1$ and $4$ has an allowed square-root radius, and any angle between $0$ and $\pi$ has an allowed half-angle.

Answer. The closed upper half-annulus $1\leq|w|\leq4,\ \Im w\geq0$ . Both circular arcs and the real-axis radial edges are included.

Check. $z=1+\mathrm{i}$ is allowed and maps to $w=2\mathrm{i}$ , inside the image.

Rotate only the source angle interval by $\pi/2$ , keeping both radius bounds unchanged.

Worked example 4: the lower half and a principal-endpoint trap

Question. Find the image of

1\leq|z|\leq2,\qquad\frac\pi2\leq\arg z\leq\pi

under $w=z^2$ .

Plan. Double the unwrapped angles first, then describe their directions geometrically.

\begin{aligned} 1\leq|w|&\leq4 &&\text{unchanged radial calculation}\\ \pi\leq2\theta&\leq2\pi &&\text{unwrapped image angles}\\ \Im w&\leq0 &&\text{closed lower-half directions} \end{aligned}

Why this step? The negative real image edge has principal argument $\pi$ , even though nearby lower-half points have principal arguments just above $-\pi$ .

Answer. The closed lower half-annulus $1\leq|w|\leq4,\ \Im w\leq0$ . In principal-angle language, it has $-\pi<\arg w\leq0$ , together with $\arg w=\pi$ .

Check. $z=-1+\mathrm{i}$ maps to $-2\mathrm{i}$ . An allowed positive-imaginary source point maps to the negative real edge, so that edge is included.

For example, cubing instead sends $0<\arg z<\pi/4$ to the open sector $0<\arg w<3\pi/4$ . The modulus is cubed; with no radial bound, all positive image radii occur.

[DIAGRAM: asset_name: A quarter-annulus squared into a half-annulus; asset_slug: square-annular-sector; caption: Squaring turns radii one and two into one and four, while doubling the source angle interval.; description: A source-plane quarter-annulus with1 at most radius at most2 and angles0 to pi/2 is paired with its image-plane upper half-annulus with radii1 and4. All circular and radial edges are solid. Each panel has independently labelled equal-scale axes, and the radius ranges and angle ranges are shown in dark ink.]
Squaring turns radii one and two into one and four, while doubling the source angle interval.

A geometric half-plane condition can express the image more cleanly than a principal interval split across the negative real axis.

Watch out

Writing only $-\pi<\arg w\leq0$ for the second image loses its negative real edge. That edge has principal argument $\pi$ . Use $\Im w\leq0$ , or add the separate condition $\arg w=\pi$ .

Shift only the angular window, preserving the same annular radii.

Invert a fractional transformation and record its missing values

For a fractional transformation, the denominator controls which source points are allowed. Solving for $z$ identifies which image values can occur.

A nonconstant fractional transformation

Let

w=\frac{az+b}{cz+d},\qquad a,b,c,d\in\mathbb C,\qquad \Delta=ad-bc\ne0

Then

(cw-a)z=b-dw,\qquad z=\frac{b-dw}{cw-a}

For $c\ne0$ , the source excludes the pole $z=-d/c$ , and the image of the full allowed plane excludes $w=a/c$ .

For $c=0$ , nonzero $\Delta$ requires $a,d\ne0$ : the map is affine, $w=(a/d)z+b/d$ , and maps the full plane bijectively onto the full plane.

Not in the formula booklet: learn it.

The nonzero- $\Delta$ condition matters. If $\Delta=0$ and the denominator is not identically zero, the map is constant wherever defined. For example,

\frac{2z+2}{z+1}=2,\qquad z\ne-1

has image $\{2\}$ , so $2$ is attained. The omitted-value rule for a nonconstant map does not apply. If $c=d=0$ , the denominator vanishes everywhere and no transformation is defined.

Worked example 5: a complex inverse

Question. For $w=(z-\mathrm{i})/(z+\mathrm{i})$ , find the inverse, its source exclusion and omitted image value. Find the image of $z=1+\mathrm{i}$ .

Plan. Rearrange before using coordinates, then substitute the source point and rationalise its denominator.

\begin{aligned} w(z+\mathrm{i})&=z-\mathrm{i} &&\text{multiply by the allowed denominator}\\ (w-1)z&=-\mathrm{i}(1+w) &&\text{collect the source variable}\\ z&=\frac{\mathrm{i}(1+w)}{1-w} &&\text{inverse for }w\ne1\\ w&=\frac1{1+2\mathrm{i}} =\frac{1-2\mathrm{i}}5 &&\text{multiply by the conjugate} \end{aligned}

Why this step? The inverse denominator reveals $w=1$ as impossible, rather than as a value to add later to an image equation.

Answer. $z=\mathrm{i}(1+w)/(1-w)$ , with source $z\ne-\mathrm{i}$ and image $w\ne1$ . The given point maps to $(1-2\mathrm{i})/5$ .

Check. Setting $w=1$ in the original relation would require $z+\mathrm{i}=z-\mathrm{i}$ , impossible for a finite $z$ .

Change only the numerator's constant from $-\mathrm{i}$ to $-2\mathrm{i}$ .

Worked example 6: the same pole, a different inverse

Question. Repeat the calculations for $w=(z-2\mathrm{i})/(z+\mathrm{i})$ .

Plan. Preserve the altered constant while collecting $z$ .

\begin{aligned} w(z+\mathrm{i})&=z-2\mathrm{i} &&\text{new numerator}\\ (w-1)z&=-\mathrm{i}(2+w) &&\text{collect terms}\\ z&=\frac{\mathrm{i}(2+w)}{1-w} &&\text{new inverse}\\ w&=\frac{1-\mathrm{i}}{1+2\mathrm{i}} =\frac{-1-3\mathrm{i}}5 &&\text{rationalise with the same conjugate} \end{aligned}

Why this step? The pole stays at the denominator's zero, while the changed numerator alters the image of every ordinary source point.

Answer. $z=\mathrm{i}(2+w)/(1-w)$ , with $z\ne-\mathrm{i}$ , $w\ne1$ ; the given point maps to $(-1-3\mathrm{i})/5$ .

Check. At $z=0$ , $w=-2$ ; the inverse then gives $z=\mathrm{i}(2-2)/(1+2)=0$ .

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FP2 4.2 - Transformations from z-plane to w-plane

Squaring can bend a straight line

Map radii and angles, then choose principal directions

Invert a fractional transformation and record its missing values

Continue this lesson.

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