Solve \(w=az+b\) for \(z\).

Substitution in both orders returns the original input, establishing both surjectivity and injectivity.

The different translation vectors show why the order matters.

Substituting \(z=i\) leaves \(R(i)=i\). Also \(|R(z)-i|=|z-i|\), since the multiplier has modulus one.

Here \(c=(3+i)/(1-i)=1+2i\). Substitution verifies \(ic+3+i=c\). Relative to c the map is multiplication by i, so \(T^4(z)=c+i^4(z-c)=z\). Checking only where zero goes would find an image point, not the center.

Write \(w=s+it\). Since \(z=1/w\), \(y=-t/(s^2+t^2)\). Hence:

In the ordinary plane remove the origin because it has no finite preimage. On the extended plane the original line contains infinity, which maps to the origin.

Here \(r^2=x^2+y^2\). Substituting the formula into \((X+iY)/(1-Z)\) gives \(x+iy\), so it really is the inverse away from the north pole.

If \(C=D\), the quadratic term vanishes. The north pole lies in the plane exactly when \(C=D\). Otherwise the equation is a circle for a genuine circular section.

Negate all three sphere coordinates and project again.

The origin and infinity are paired separately. Equatorial antipodes both have modulus one; they are not an inside/outside pair.

Write \(1/z=(x-iy)/r^2\). Substitute its coordinates into the inverse projection and multiply numerator and denominator by \(r^2\). The result is \((2x,-2y,1-r^2)/(1+r^2)\), exactly the stated half-turn. The origin and north pole are exchanged, so the formula extends to those two points as well.

Approach. Plot the images of 0, 1, and i first, then transform several grid lines.

• a. Send each grid point \((x,y)\) to \((x+2,y)\): the whole grid moves two units right.
• b. Send \((x,y)\) to \((-y,x)\): for example \(1\mapsto i\) and \(i\mapsto-1\).
• c. Send \((x,y)\) to \((5x,5y)\); every distance from the origin is multiplied by \(5\).
• d. Send \((x,y)\) to \((x-3,y+1)\): three units left and one up.

Approach. Compare with z+b, e^(iθ)z, and kz.

• a. Translation by \(2\).
• b. Counterclockwise rotation about the origin through \(\pi/2\).
• c. Homothety about the origin with positive scale \(5\).
• d. Translation by \(-3+i\).

Approach. The circle’s radius is inversely proportional to the original line’s distance from zero.

A horizontal line \(y=k\ne0\) maps to a circle of radius \(1/(2|k|)\); a vertical line \(x=k\ne0\) has the same radius formula. Consequently, among the displayed lines, those farthest from the origin give the smallest circles. Lines equally far away tie. The coordinate axes themselves are exceptional: their extended images are lines, not finite-radius circles.

Approach. Substitute the line’s height into the completed-square formula.

Here \(k=1\). Substitution into the formula gives \(s^2+(t+1/2)^2=1/4\): center \(-i/2\), radius \(1/2\). In the ordinary plane the image omits zero; when the original line includes infinity, its extended image includes zero.

Approach. Find each circle’s center, then take the dot product of its radius vectors at the intersection.

Writing \(w=u+iv\), the horizontal line gives \(u^2+(v+1/(2b))^2=1/(4b^2)\); the vertical line gives \((u-1/(2a))^2+v^2=1/(4a^2)\). Their original intersection \(a+ib\) maps to \(w_0=(a-ib)/(a^2+b^2)\). At this point, the two radius vectors are \((u,v+1/(2b))\) and \((u-1/(2a),v)\). Their dot product is \(u^2+v^2-u/(2a)+v/(2b)=0\), using \(u=a/(a^2+b^2),v=-b/(a^2+b^2)\). Perpendicular radii give perpendicular tangents, so the right angle is preserved. If one original line is an axis, its image remains an axis and the same right-angle conclusion holds at any nonzero intersection.

Approach. The denominator is positive everywhere except the north pole.

For \(P=(a,b,c)\ne(0,0,1)\), the projection is \((a+ib)/(1-c)\). Since \(1-c>0\), the signs of the planar coordinates equal those of \(a,b\). The open first quadrant corresponds to \(a>0,b>0\) on the sphere, a spherical lune between the two coordinate meridians. Include the meridian edges if “first quadrant” includes the axes; the north pole always corresponds to infinity rather than a finite point.

Approach. Interpret the axes as circles through the north pole on the sphere.

Yes. On either axis there are sequences whose moduli tend to infinity; on the sphere both axes close up into circles through the north pole. The same single point at infinity belongs to both extended axes. It is not an ordinary finite point of either axis.

Approach. Follow a line indefinitely in either direction.

Every straight line. A line can be parametrized \(z=a+tv\) with \(v\ne0,t\in\mathbb R\). The estimate \(|a+tv|\ge |t||v|-|a|\) shows that the modulus tends to infinity as \(|t|\to\infty\). Its closure on the Riemann sphere therefore includes the same point \(\infty\).

Approach. Compute the inner function first.

Substitution gives \(f(g(z))=i(-iz+5)+2=z+2+5i\). This is a translation two units right and five units up. To sketch it, draw an arrow from any sample \((x,y)\) to \((x+2,y+5)\); every arrow has the same displacement.

Approach. Write each transformation in its standard formula.

• Translations. \((z+d)+b=z+(d+b)\).
• Homotheties. For \(k,l>0\), \(k(lz)=(kl)z\) and \(kl>0\).
• Rotations. \(e^{i\theta}(e^{i\phi}z)=e^{i(\theta+\phi)}z\).

The common-center hypothesis matters. Rotations with different centers can compose to a nonzero translation, so the unrestricted claim about arbitrary rotations would be false.

Approach. Use the polar form of a.

Write \(a=re^{i\theta}\) with \(r=|a|>0\). The composition \(z\mapsto e^{i\theta}z\mapsto re^{i\theta}z\mapsto re^{i\theta}z+b\) equals \(az+b\). If \(a=0\), the function is constant, not bijective, and cannot be a composition of these transformations.

Approach. Follow the same pair of tangent directions through both maps.

Let two regular curves meet at \(p\). If \(f\) is conformal there, their images meet at \(f(p)\) with the original angle. If \(g\) is conformal at \(f(p)\), the images under \(g\) retain that same angle. Thus \(g\circ f\) preserves the angle of every such pair. Analytically, for holomorphic maps the chain rule gives \((g\circ f)^{\prime}(p)=g^{\prime}(f(p))f^{\prime}(p)\ne0\), also proving local conformality.

Approach. Conjugate an origin-centered rotation by a translation.

A rotation by \(\theta\) about \(c\) is \(c+e^{i\theta}(z-c)\): translate the center to zero, rotate, and translate back.

• a. \(i+\frac{1+i}{\sqrt2}(z-i)\)
• b. \(2+i+i(z-2-i)=iz+3-i\)
• c. \(3+(\frac12-\frac{\sqrt3}{2}i)(z-3)\)

Approach. Subtract the center before scaling, then add it back.

A homothety of scale \(k>0\) and center \(c\) is \(c+k(z-c)\). Hence (a) \(i+3(z-i)=3z-2i\); (b) \(-2+(z+2)/2=z/2-1\). Each fixes its center and multiplies displacement from it by the stated scale.

Approach. A rotation before the same scale and translation supplies a second map.

Use \(f(z)=3z+5\) and \(g(z)=3iz+5\). For either map, the distance of the image from \(5\) is \(3|z|\). Thus the image lies in the target disk exactly when \(|z|<1\). Conversely, for a target point \(w\), the preimages \((w-5)/3\) and \((w-5)/(3i)\) lie in the unit disk. This proves onto, not merely into.

Approach. Rationalize each reciprocal, then use the original line equation.

The images are \(1,(1-i)/2,(1+i)/2,(1-2i)/5\). Every original point satisfies \(\operatorname{Re}z=1\). If \(w=u+iv=1/z\), then \(\operatorname{Re}(1/w)=u/(u^2+v^2)=1\). Hence \((u-1/2)^2+v^2=1/4\). The four computed points satisfy this circle equation. The image of the entire finite line is this circle with \(0\) removed.

Approach. Simplify the quotient of two image secants before taking its argument.

Because \(p/q\) is positive real, the two ratios have the same argument, proving equality of the directed secant angles. Taking regular-curve limits at a nonzero \(z\) gives equality of tangent angles. If \(p,q\) are merely on the same line but opposite rays, the multiplier is negative and arguments differ by \(\pi\); near a fixed nonzero \(z\) one can choose the same-ray situation needed for the limiting proof. Alternatively the derivative \(-1/z^2\ne0\) directly proves local conformality.

Approach. Compute the imaginary parts and the squared moduli from stereographic projection.

Let \(P=(a,b,c)\) and \(-P=(-a,-b,-c)\). Away from the poles their images are \(z=(a+ib)/(1-c)\) and \(w=(-a-ib)/(1+c)\). Their imaginary parts have opposite signs when \(b\ne0\); if \(b=0\), both are real. Their squared moduli are \(|z|^2=(1+c)/(1-c)\) and its reciprocal. If \(c\ne0\), one lies inside and the other outside the unit circle; if \(c=0\), both lie on the circle. The poles map to \(0\) and \(\infty\).

Approach. Use the unit-sphere equation to simplify the product.

With the notation of Exercise 10 and excluding the poles,

The final equality uses \(a^2+b^2+c^2=1\). Equivalently \(w=-1/\bar z\). At the two poles use the extended convention \(0\leftrightarrow\infty\); the product expression is not finite arithmetic there.

Approach. Eliminate a and b using the projection equation, then solve for c.

Set \(r^2=x^2+y^2\). Then \(a=x(1-c),b=y(1-c)\), so \(r^2(1-c)^2=1-c^2\). Since the finite image excludes \(c=1\), divide by \(1-c\) to get \(r^2(1-c)=1+c\). Solving yields

Direct substitution verifies both the sphere equation and the original projection. Infinity maps to \((0,0,1)\).

Approach. Apply stereographic projection after the coordinate rotation.

The sphere rotation sends \((a,b,c)\) to \((-a,b,-c)\). Its projected image is \((-a+ib)/(1+c)\). For \(z=(a+ib)/(1-c)\), rationalization and \(a^2+b^2=1-c^2\) give \(-1/z=(-a+ib)/(1+c)\). Thus the plane map is \(z\mapsto-1/z\), extended by exchanging zero and infinity.

Approach. Use u=x²−y² and v=2xy; eliminate the free coordinate in each line.

Put \(w=u+iv=(x^2-y^2)+2xyi\). For \(y=b\ne0\), eliminating \(x=v/(2b)\) gives \(u=v^2/(4b^2)-b^2\). For \(x=a\ne0\), eliminating \(y=v/(2a)\) gives \(u=a^2-v^2/(4a^2)\). Equating these yields \(v=\pm2ab,u=a^2-b^2\), two distinct intersections. At them the products of the two derivatives \(du/dv\) are \([v/(2b^2)][-v/(2a^2)]=-1\), so their tangent vectors are perpendicular. One intersection is the image of the original meeting point \(a+ib\); the other occurs because the distinct points \(-a+ib\) and \(a-ib\) are negatives and have equal squares. If \(a=0\) or \(b=0\), one parabola degenerates to a ray and the two-intersection assertion no longer applies. At zero, squaring is not conformal because its derivative vanishes.

Approach. Compare degrees, but check whether the denominator actually depends on z.

• a. The denominator’s modulus tends to infinity, so the limit is \(0\).
• b. Divide by \(z\): \((2-1/z)/(1+i/z)\to2\).
• c. Divide by \(z\): \((2-i/z)/(i+i/z)\to2/i=-2i\).
• d. The denominator is a fixed nonzero constant and the numerator’s modulus tends to infinity, so the spherical limit is \(\infty\).

We have \(R_b(R_a(z))=2b-(2a-z)=z+2(b-a)\), whereas \(R_a(R_b(z))=z+2(a-b)\). These are opposite translations. Since the centers are distinct, neither displacement is zero and the maps differ. Each individual half-turn has a fixed center; their composition has no finite fixed point.

Substitution yields \(2x=x^2+y^2-1\), or \((x-1)^2+y^2=2\). The sphere plane passes through the origin, so the section is a great circle. A finite point satisfies \(|z|^2-2\operatorname{Re}z-1=0\), which excludes z=0. For \(w=-1/\bar z\), multiply \(|w|^2-2\operatorname{Re}w-1\) by \(|z|^2\): the result is \(1+2\operatorname{Re}z-|z|^2=0\). Hence the antipodal partner lies on the same circle, as the spherical picture predicts.