Complex Numbers: Mixed Practice, Worked Solutions

Ques­tion 1: A neg­a­tive power of a com­plex num­ber#

The prob­lem#

Express $(1 - 2i)^{-3}$ in the form $a + ib$.

Under­stand­ing the prob­lem#

A neg­a­tive power means "one over the pos­i­tive power":

$$\displaystyle (1 - 2i)^{-3} = \frac{1}{(1 - 2i)^3}.$$

So you are asked for the rec­i­p­ro­cal of $(1-2i)^3$, writ­ten with a real part and an imag­i­nary part.

The idea#

First expand $(1 - 2i)^3$ using $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ and the pow­ers $i^2 = -1$, $i^3 = -i$. Then find the rec­i­p­ro­cal by mul­ti­ply­ing numer­a­tor and denom­i­na­tor by the con­ju­gate.

Step-by-step solu­tion#

Step 1. Expand the cube. Take $a = 1$ and $b = 2i$.

$$\begin{aligned} (1 - 2i)^3 &= 1^3 - 3(1)^2(2i) + 3(1)(2i)^2 - (2i)^3 \\ &= 1 - 6i + 3(4i^2) - 8i^3 \\ &= 1 - 6i + 3(-4) - 8(-i) \\ &= 1 - 6i - 12 + 8i \\ &= -11 + 2i. \end{aligned}$$

Step 2. Write the rec­i­p­ro­cal and mul­ti­ply top and bot­tom by the con­ju­gate $-11 - 2i$, so the denom­i­na­tor becomes real.

$$\displaystyle \frac{1}{-11 + 2i} = \frac{-11 - 2i}{(-11 + 2i)(-11 - 2i)}.$$

Step 3. The denom­i­na­tor is of the form $(p + iq)(p - iq) = p^2 + q^2$.

$$(-11)^2 + 2^2 = 121 + 4 = 125.$$

Step 4. Split into real and imag­i­nary parts.

$$\displaystyle (1 - 2i)^{-3} = \frac{-11 - 2i}{125} = -\frac{11}{125} - \frac{2}{125}\,i.$$

Check­ing the answer#

Mul­ti­ply back: $\displaystyle (-11 + 2i)\cdot\frac{-11 - 2i}{125} = \frac{121 + 4}{125} = 1$. Cor­rect.

Answer#

$\displaystyle (1 - 2i)^{-3} = -\frac{11}{125} - \frac{2}{125}\,i$.

Com­mon mis­take to avoid#

Do not write $(1-2i)^{-3} = 1^{-3} - (2i)^{-3}$. Pow­ers do not split over a sum.

Ques­tion 2: Find­ing $x$ and $y$ from a prod­uct#

The prob­lem#

Find $x, y$ if $(x + iy)(2 - 3i) = 4 + i$.

Under­stand­ing the prob­lem#

Here $x$ and $y$ are real num­bers. The unknown com­plex num­ber $x + iy$, mul­ti­plied by $2 - 3i$, gives $4 + i$. You need the real part $x$ and the imag­i­nary part $y$.

The idea#

Divide both sides by $2 - 3i$, then sim­plify the quo­tient to $a + ib$ using the con­ju­gate $2 + 3i$. Two com­plex num­bers are equal only when their real parts and imag­i­nary parts are equal, so you can read off $x$ and $y$.

Step-by-step solu­tion#

Step 1. Iso­late $x + iy$.

$$\displaystyle x + iy = \frac{4 + i}{2 - 3i}.$$

Step 2. Mul­ti­ply top and bot­tom by the con­ju­gate $2 + 3i$.

$$\displaystyle x + iy = \frac{(4 + i)(2 + 3i)}{(2 - 3i)(2 + 3i)}.$$

Step 3. Expand the numer­a­tor.

$$\begin{aligned} (4 + i)(2 + 3i) &= 8 + 12i + 2i + 3i^2 \\ &= 8 + 14i - 3 \\ &= 5 + 14i. \end{aligned}$$

Step 4. The denom­i­na­tor is $2^2 + 3^2 = 13$.

$$\displaystyle x + iy = \frac{5 + 14i}{13} = \frac{5}{13} + \frac{14}{13}\,i.$$

Step 5. Com­pare real and imag­i­nary parts.

$$\displaystyle x = \frac{5}{13}, \qquad y = \frac{14}{13}.$$

Check­ing the answer#

$\displaystyle \left(\frac{5 + 14i}{13}\right)(2 - 3i) = \frac{10 - 15i + 28i - 42i^2}{13} = \frac{52 + 13i}{13} = 4 + i$. Cor­rect.

Answer#

$\displaystyle x = \frac{5}{13}$, $\displaystyle y = \frac{14}{13}$.

Ques­tion 3: Sum of pow­ers of $i$#

The prob­lem#

Eval­u­ate $i^{17} + i^{22} + i^{35} + i^{100}$.

Under­stand­ing the prob­lem#

You need the value of each power of $i$, then their sum.

The idea#

The pow­ers of $i$ repeat in a cycle of four: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$. So $i^n$ depends only on the remain­der when $n$ is divided by $4$.

Step-by-step solu­tion#

Step 1. Divide each expo­nent by $4$ and note the remain­der.

$$\begin{aligned} 17 &= 4 \times 4 + 1 &&\Rightarrow\; i^{17} = i^1 = i \\ 22 &= 4 \times 5 + 2 &&\Rightarrow\; i^{22} = i^2 = -1 \\ 35 &= 4 \times 8 + 3 &&\Rightarrow\; i^{35} = i^3 = -i \\ 100 &= 4 \times 25 + 0 &&\Rightarrow\; i^{100} = i^0 = 1 \end{aligned}$$

Step 2. Add.

$$i + (-1) + (-i) + 1 = 0.$$

Check­ing the answer#

Four con­sec­u­tive-type val­ues $i, -1, -i, 1$ (one from each posi­tion of the cycle) always add to $0$, which agrees.

Answer#

$i^{17} + i^{22} + i^{35} + i^{100} = 0$.

Ques­tion 4: Mul­ti­plica­tive inverse#

The prob­lem#

Find the mul­ti­plica­tive inverse of $\sqrt{5} + 3i$.

Under­stand­ing the prob­lem#

The mul­ti­plica­tive inverse of $z$ is the num­ber $w$ with $zw = 1$, that is $\displaystyle w = \frac{1}{z}$. You must write it as $a + ib$.

The idea#

For $z = a + ib$, $\displaystyle \frac{1}{z} = \frac{\bar z}{\lvert z \rvert^2} = \frac{a - ib}{a^2 + b^2}$.

Step-by-step solu­tion#

Step 1. Write the con­ju­gate: $\bar z = \sqrt{5} - 3i$.

Step 2. Find $\lvert z \rvert^2$.

$$\lvert z \rvert^2 = (\sqrt{5})^2 + 3^2 = 5 + 9 = 14.$$

Step 3. Divide.

$$\displaystyle \frac{1}{\sqrt{5} + 3i} = \frac{\sqrt{5} - 3i}{14} = \frac{\sqrt{5}}{14} - \frac{3}{14}\,i.$$

Check­ing the answer#

$(\sqrt{5} + 3i)(\sqrt{5} - 3i) = 5 + 9 = 14$, and $14 / 14 = 1$. Cor­rect.

Answer#

The inverse is $\displaystyle \frac{\sqrt{5}}{14} - \frac{3}{14}\,i$.

Ques­tion 5: Prov­ing $x^2 + y^2 = 1$#

The prob­lem#

If $\displaystyle x + iy = \frac{a + ib}{a - ib}$, prove $x^2 + y^2 = 1$.

Under­stand­ing the prob­lem#

Here $a, b$ are real and not both zero, so $a - ib \neq 0$. Notice that $x^2 + y^2$ is exactly $\lvert x + iy \rvert^2$. So you must show that the mod­u­lus of the given quo­tient is $1$.

The idea#

Use two facts from the les­son: the mod­u­lus of a quo­tient is the quo­tient of the mod­uli, and a num­ber and its con­ju­gate have the same mod­u­lus.

Step-by-step solu­tion#

Step 1. Take the mod­u­lus of both sides.

$$\displaystyle \lvert x + iy \rvert = \left\lvert \frac{a + ib}{a - ib} \right\rvert = \frac{\lvert a + ib \rvert}{\lvert a - ib \rvert}.$$

Step 2. Com­pute each mod­u­lus.

$$\lvert a + ib \rvert = \sqrt{a^2 + b^2}, \qquad \lvert a - ib \rvert = \sqrt{a^2 + (-b)^2} = \sqrt{a^2 + b^2}.$$

Step 3. So the quo­tient of mod­uli is $1$.

$$\displaystyle \lvert x + iy \rvert = \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}} = 1.$$

Step 4. Square both sides. Since $\lvert x + iy \rvert^2 = x^2 + y^2$,

$$x^2 + y^2 = 1. \qquad \blacksquare$$

Check­ing the answer#

Try $a = 1, b = 1$: $\displaystyle \frac{1 + i}{1 - i} = \frac{(1+i)^2}{2} = \frac{2i}{2} = i$, so $x = 0, y = 1$ and $x^2 + y^2 = 1$. Cor­rect.

Answer#

$x^2 + y^2 = \lvert x + iy \rvert^2 = 1$, as required.

Ques­tion 6: Mod­u­lus of a quo­tient#

The prob­lem#

Find the mod­u­lus of $\displaystyle \frac{(3 + 4i)(5 - 12i)}{6 + 8i}$.

Under­stand­ing the prob­lem#

You want one real, non-neg­a­tive num­ber: the mod­u­lus of the whole expres­sion.

The idea#

You do not need to mul­ti­ply out. Use $\lvert z_1 z_2 \rvert = \lvert z_1 \rvert\,\lvert z_2 \rvert$ and $\displaystyle \left\lvert \frac{z_1}{z_2} \right\rvert = \frac{\lvert z_1 \rvert}{\lvert z_2 \rvert}$.

Step-by-step solu­tion#

Step 1. Find each mod­u­lus.

$$\begin{aligned} \lvert 3 + 4i \rvert &= \sqrt{9 + 16} = 5 \\ \lvert 5 - 12i \rvert &= \sqrt{25 + 144} = 13 \\ \lvert 6 + 8i \rvert &= \sqrt{36 + 64} = 10 \end{aligned}$$

Step 2. Com­bine.

$$\displaystyle \left\lvert \frac{(3 + 4i)(5 - 12i)}{6 + 8i} \right\rvert = \frac{5 \times 13}{10} = \frac{65}{10} = \frac{13}{2}.$$

Check­ing the answer#

Notice $6 + 8i = 2(3 + 4i)$, so the expres­sion is $\displaystyle \frac{5 - 12i}{2}$, whose mod­u­lus is $\displaystyle \frac{13}{2}$. The same answer by a sec­ond route.

Answer#

The mod­u­lus is $\displaystyle \frac{13}{2}$.

Ques­tion 7: Polar forms#

The prob­lem#

Con­vert to polar form: (a) $-2 + 2\sqrt{3}i$; (b) $\displaystyle \frac{1 - i}{\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}}$.

Under­stand­ing the prob­lem#

Polar form is $r(\cos\theta + i\sin\theta)$, where $r = \lvert z \rvert$ and $\theta$ is the prin­ci­pal argu­ment, $-\pi < \theta \le \pi$.

The idea#

Find $r$ from $\sqrt{a^2 + b^2}$. Find the angle from $\displaystyle \tan\alpha = \left\lvert \frac{b}{a} \right\rvert$, then place it in the right quad­rant using the signs of $a$ and $b$. For a quo­tient, divide the mod­uli and sub­tract the argu­ments.

Step-by-step solu­tion#

Part (a)

Step 1. Mod­u­lus.

$$r = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = 4.$$

Step 2. Ref­er­ence angle.

$$\displaystyle \tan\alpha = \frac{2\sqrt{3}}{2} = \sqrt{3} \;\Rightarrow\; \alpha = \frac{\pi}{3}.$$

Step 3. Quad­rant. The real part is neg­a­tive and the imag­i­nary part is pos­i­tive, so the point lies in the sec­ond quad­rant.

$$\displaystyle \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}.$$

Step 4. Write the polar form.

$$\displaystyle -2 + 2\sqrt{3}i = 4\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right).$$

Part (b)

Step 1. Put the numer­a­tor in polar form. $\lvert 1 - i \rvert = \sqrt{2}$, and $1 - i$ lies in the fourth quad­rant with ref­er­ence angle $\displaystyle \frac{\pi}{4}$, so its argu­ment is $\displaystyle -\frac{\pi}{4}$.

$$\displaystyle 1 - i = \sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right).$$

Step 2. The denom­i­na­tor is already in polar form, with mod­u­lus $1$ and argu­ment $\displaystyle \frac{\pi}{3}$.

Step 3. Divide the mod­uli and sub­tract the argu­ments.

$$\displaystyle \begin{aligned} r &= \frac{\sqrt{2}}{1} = \sqrt{2}, \\ \theta &= -\frac{\pi}{4} - \frac{\pi}{3} = -\frac{3\pi}{12} - \frac{4\pi}{12} = -\frac{7\pi}{12}. \end{aligned}$$

Step 4. $\displaystyle -\frac{7\pi}{12}$ lies between $-\pi$ and $\pi$, so it is already the prin­ci­pal argu­ment.

$$\displaystyle \frac{1 - i}{\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}} = \sqrt{2}\left(\cos\left(-\frac{7\pi}{12}\right) + i\sin\left(-\frac{7\pi}{12}\right)\right).$$

Check­ing the answer#

(a) $\displaystyle 4\cos\frac{2\pi}{3} = 4 \times \left(-\frac12\right) = -2$ and $\displaystyle 4\sin\frac{2\pi}{3} = 4 \times \frac{\sqrt3}{2} = 2\sqrt3$. Cor­rect.
(b) $\displaystyle -\frac{7\pi}{12}$ is about $-105^\circ$, in the third quad­rant. Directly, $\displaystyle \frac{1 - i}{\frac12 + \frac{\sqrt3}{2}i}$ has neg­a­tive real and neg­a­tive imag­i­nary parts, which agrees.

Answer#

(a) $\displaystyle 4\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$; (b) $\displaystyle \sqrt{2}\left(\cos\left(-\frac{7\pi}{12}\right) + i\sin\left(-\frac{7\pi}{12}\right)\right)$.

Com­mon mis­take to avoid#

In (a), $\displaystyle \tan^{-1}\left(\frac{2\sqrt3}{-2}\right)$ on a cal­cu­la­tor gives $\displaystyle -\frac{\pi}{3}$, which is in the wrong quad­rant. Always check the signs of the real and imag­i­nary parts.

Ques­tion 8: Qua­drat­ics with com­plex roots#

The prob­lem#

Solve: (a) $5x^2 - 6x + 2 = 0$; (b) $27x^2 - 10x + 1 = 0$; (c) $x^2 - x + 2 = 0$.

Under­stand­ing the prob­lem#

Each is a qua­dratic $ax^2 + bx + c = 0$ with real coef­fi­cients. When the dis­crim­i­nant is neg­a­tive, the roots are a pair of com­plex con­ju­gates.

The idea#

Use the qua­dratic for­mula

$$\displaystyle x = \frac{-b \pm \sqrt{D}}{2a}, \qquad D = b^2 - 4ac,$$

and when $D < 0$ write $\sqrt{D} = \sqrt{-D}\,i$.

Step-by-step solu­tion#

Part (a)

Step 1. Here $a = 5$, $b = -6$, $c = 2$.

$$D = (-6)^2 - 4(5)(2) = 36 - 40 = -4, \qquad \sqrt{D} = 2i.$$

Step 2. Apply the for­mula.

$$\displaystyle x = \frac{6 \pm 2i}{10} = \frac{3 \pm i}{5}.$$

Part (b)

Step 1. Here $a = 27$, $b = -10$, $c = 1$.

$$D = 100 - 108 = -8, \qquad \sqrt{D} = \sqrt{8}\,i = 2\sqrt{2}\,i.$$

Step 2. Apply the for­mula and can­cel the com­mon fac­tor $2$.

$$\displaystyle x = \frac{10 \pm 2\sqrt{2}\,i}{54} = \frac{5 \pm \sqrt{2}\,i}{27}.$$

Part (c)

Step 1. Here $a = 1$, $b = -1$, $c = 2$.

$$D = 1 - 8 = -7, \qquad \sqrt{D} = \sqrt{7}\,i.$$

Step 2. Apply the for­mula.

$$\displaystyle x = \frac{1 \pm \sqrt{7}\,i}{2}.$$

Check­ing the answer#

Use sum and prod­uct of roots. In (a) the roots add to $\displaystyle \frac{6}{5} = -\frac{b}{a}$ and mul­ti­ply to $\displaystyle \frac{9 + 1}{25} = \frac{2}{5} = \frac{c}{a}$. In (b) the sum is $\displaystyle \frac{10}{27}$ and the prod­uct is $\displaystyle \frac{25 + 2}{729} = \frac{1}{27}$. In (c) the sum is $1$ and the prod­uct is $\displaystyle \frac{1 + 7}{4} = 2$. All agree.

Answer#

(a) $\displaystyle \frac{3 \pm i}{5}$; (b) $\displaystyle \frac{5 \pm \sqrt{2}\,i}{27}$; (c) $\displaystyle \frac{1 \pm \sqrt{7}\,i}{2}$.

Ques­tion 9: Roots of $x^2 + x + 1 = 0$#

The prob­lem#

If $\alpha$ and $\beta$ are the roots of $x^2 + x + 1 = 0$, find $\alpha^2 + \beta^2$ and show $\alpha^3 = 1$.

Under­stand­ing the prob­lem#

There are two tasks: a value to com­pute and a state­ment to prove. You do not need to find $\alpha$ and $\beta$ explic­itly.

The idea#

For $ax^2 + bx + c = 0$, the sum of roots is $\displaystyle -\frac{b}{a}$ and the prod­uct is $\displaystyle \frac{c}{a}$. For the sec­ond part, use the fac­tori­sa­tion $\alpha^3 - 1 = (\alpha - 1)(\alpha^2 + \alpha + 1)$.

Step-by-step solu­tion#

Step 1. Sum and prod­uct of the roots ($a = b = c = 1$).

$$\alpha + \beta = -1, \qquad \alpha\beta = 1.$$

Step 2. Use the iden­tity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.

$$\alpha^2 + \beta^2 = (-1)^2 - 2(1) = 1 - 2 = -1.$$

Step 3. Since $\alpha$ is a root, it sat­is­fies the equa­tion.

$$\alpha^2 + \alpha + 1 = 0.$$

Step 4. Fac­torise $\alpha^3 - 1$ and sub­sti­tute.

$$\alpha^3 - 1 = (\alpha - 1)(\alpha^2 + \alpha + 1) = (\alpha - 1) \times 0 = 0.$$

So $\alpha^3 = 1$. $\blacksquare$

Check­ing the answer#

The roots are $\displaystyle \frac{-1 \pm \sqrt{3}\,i}{2}$. For $\displaystyle \alpha = \frac{-1 + \sqrt3 i}{2}$, $\displaystyle \alpha^2 = \frac{1 - 2\sqrt3 i - 3}{4} = \frac{-1 - \sqrt3 i}{2} = \beta$, and $\alpha^2 + \beta^2 = \beta + \alpha = -1$. Cor­rect.

Answer#

$\alpha^2 + \beta^2 = -1$, and $\alpha^3 = 1$ because $\alpha^3 - 1 = (\alpha - 1)(\alpha^2 + \alpha + 1) = 0$.

Ques­tion 10: When are the roots non-real?#

The prob­lem#

For which real $k$ does $x^2 + kx + 9 = 0$ have non-real roots?

Under­stand­ing the prob­lem#

"Non-real" means the roots are com­plex with a non-zero imag­i­nary part. You need the set of val­ues of $k$.

The idea#

A qua­dratic with real coef­fi­cients has non-real roots exactly when its dis­crim­i­nant is neg­a­tive, $D < 0$.

Step-by-step solu­tion#

Step 1. Here $a = 1$, $b = k$, $c = 9$.

$$D = k^2 - 4(1)(9) = k^2 - 36.$$

Step 2. Require $D < 0$.

$$k^2 - 36 < 0 \;\Longleftrightarrow\; k^2 < 36.$$

Step 3. A square is less than $36$ exactly when the num­ber lies strictly between $-6$ and $6$.

$$-6 < k < 6.$$

Check­ing the answer#

$k = 0$ gives $x^2 + 9 = 0$, roots $\pm 3i$: non-real. $k = 6$ gives $(x + 3)^2 = 0$, a real repeated root, so $6$ is cor­rectly excluded.

Answer#

$-6 < k < 6$.

Com­mon mis­take to avoid#

Do not write $k < 6$ only. Tak­ing square roots of $k^2 < 36$ gives $\lvert k \rvert < 6$, which has two sides.

Ques­tion 11: Cubes of two con­ju­gate quo­tients#

The prob­lem#

If $\displaystyle \left(\frac{1 + i}{1 - i}\right)^3 - \left(\frac{1 - i}{1 + i}\right)^3 = x + iy$, find $(x, y)$.

Under­stand­ing the prob­lem#

You must sim­plify the left side to the form $x + iy$ and read off the real num­bers $x$ and $y$.

The idea#

Sim­plify each quo­tient first. It turns out to be a sim­ple power of $i$, and then cub­ing is easy.

Step-by-step solu­tion#

Step 1. Sim­plify the first quo­tient by mul­ti­ply­ing by the con­ju­gate $1 + i$.

$$\displaystyle \frac{1 + i}{1 - i} = \frac{(1 + i)^2}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{2} = \frac{2i}{2} = i.$$

Step 2. The sec­ond quo­tient is the rec­i­p­ro­cal of the first.

$$\displaystyle \frac{1 - i}{1 + i} = \frac{1}{i} = \frac{-i}{-i \cdot i} = \frac{-i}{1} = -i.$$

Step 3. Cube each.

$$i^3 = -i, \qquad (-i)^3 = -i^3 = -(-i) = i.$$

Step 4. Sub­tract.

$$(-i) - (i) = -2i = 0 + (-2)i.$$

Step 5. Com­pare with $x + iy$.

$$x = 0, \qquad y = -2.$$

Check­ing the answer#

$\displaystyle \left(\frac{1-i}{1+i}\right)^3$ is the con­ju­gate of $\displaystyle \left(\frac{1+i}{1-i}\right)^3$. A num­ber minus its con­ju­gate is $2i \times$ (imag­i­nary part), so the result must be purely imag­i­nary. It is.

Answer#

$(x, y) = (0, -2)$.

Ques­tion 12: A mod­u­lus equal to $1$#

The prob­lem#

If $z_1, z_2$ are com­plex with $\lvert z_1 \rvert = \lvert z_2 \rvert = 1$, show $\displaystyle \left\lvert \frac{z_1 - z_2}{1 - \bar z_1 z_2} \right\rvert = 1$ when­ever the denom­i­na­tor is non-zero. (Hint: $\lvert w \rvert^2 = w\bar w$.)

Under­stand­ing the prob­lem#

Both $z_1$ and $z_2$ lie on the unit cir­cle, so $z_1\bar z_1 = 1$ and $z_2\bar z_2 = 1$. You must show that the numer­a­tor and the denom­i­na­tor have the same mod­u­lus; then their quo­tient has mod­u­lus $1$.

The idea#

It is eas­ier to com­pare squares of mod­uli. Expand $\lvert z_1 - z_2 \rvert^2$ and $\lvert 1 - \bar z_1 z_2 \rvert^2$ with $\lvert w \rvert^2 = w\bar w$, and show they are equal.

Step-by-step solu­tion#

Step 1. Write down what the given con­di­tion means.

$$z_1\bar z_1 = \lvert z_1 \rvert^2 = 1, \qquad z_2\bar z_2 = \lvert z_2 \rvert^2 = 1.$$

Step 2. Expand the square of the numer­a­tor's mod­u­lus. The con­ju­gate of $z_1 - z_2$ is $\bar z_1 - \bar z_2$.

$$\begin{aligned} \lvert z_1 - z_2 \rvert^2 &= (z_1 - z_2)(\bar z_1 - \bar z_2) \\ &= z_1\bar z_1 - z_1\bar z_2 - z_2\bar z_1 + z_2\bar z_2 \\ &= 1 - z_1\bar z_2 - \bar z_1 z_2 + 1 \\ &= 2 - z_1\bar z_2 - \bar z_1 z_2. \end{aligned}$$

Step 3. Expand the square of the denom­i­na­tor's mod­u­lus. The con­ju­gate of $1 - \bar z_1 z_2$ is $1 - z_1\bar z_2$.

$$\begin{aligned} \lvert 1 - \bar z_1 z_2 \rvert^2 &= (1 - \bar z_1 z_2)(1 - z_1\bar z_2) \\ &= 1 - z_1\bar z_2 - \bar z_1 z_2 + (\bar z_1 z_1)(z_2\bar z_2) \\ &= 1 - z_1\bar z_2 - \bar z_1 z_2 + (1)(1) \\ &= 2 - z_1\bar z_2 - \bar z_1 z_2. \end{aligned}$$

Step 4. The two results are iden­ti­cal, so

$$\lvert z_1 - z_2 \rvert^2 = \lvert 1 - \bar z_1 z_2 \rvert^2 \;\Rightarrow\; \lvert z_1 - z_2 \rvert = \lvert 1 - \bar z_1 z_2 \rvert,$$

since both mod­uli are non-neg­a­tive.

Step 5. The denom­i­na­tor is non-zero, so we may divide.

$$\displaystyle \left\lvert \frac{z_1 - z_2}{1 - \bar z_1 z_2} \right\rvert = \frac{\lvert z_1 - z_2 \rvert}{\lvert 1 - \bar z_1 z_2 \rvert} = 1. \qquad \blacksquare$$

Check­ing the answer#

Take $z_1 = 1$, $z_2 = i$: $\displaystyle \frac{1 - i}{1 - i} = 1$, whose mod­u­lus is $1$. Take $z_1 = i$, $z_2 = -1$: $\displaystyle \frac{i + 1}{1 - (-i)(-1)} = \frac{1 + i}{1 - i} = i$, mod­u­lus $1$. Both agree.

Answer#

$\lvert z_1 - z_2 \rvert^2 = \lvert 1 - \bar z_1 z_2 \rvert^2 = 2 - z_1\bar z_2 - \bar z_1 z_2$, so the quo­tient has mod­u­lus $1$.