Complex Numbers and Their Algebra, Worked Solutions

Ques­tion 1: Addi­tion, mul­ti­pli­ca­tion and pow­ers of i#

The prob­lem#

Write in the form $a + ib$: (a) $(5 - 3i) + (-2 + 7i)$; (b) $(4 + i)(2 - 3i)$; (c) $i^{27}$; (d) $i^{-18}$; (e) $(2 - i)^3$.

Under­stand­ing the prob­lem#

Each expres­sion is a com­plex num­ber writ­ten in an untidy way. You must sim­plify each one until it looks like (real num­ber) $+ i\,$(real num­ber). The only spe­cial fact you need is $i^2 = -1$.

The idea#

Add by col­lect­ing real parts and imag­i­nary parts sep­a­rately. Mul­ti­ply by expand­ing the brack­ets and replac­ing $i^2$ by $-1$. For pow­ers of $i$, use the cycle of four: $i^{4k} = 1$, $i^{4k+1} = i$, $i^{4k+2} = -1$, $i^{4k+3} = -i$.

Step-by-step solu­tion#

Part (a)

Step 1. Group the real parts together and the imag­i­nary parts together.

$$(5 - 3i) + (-2 + 7i) = (5 - 2) + i(-3 + 7)$$

Step 2. Sim­plify each group.

$$= 3 + 4i$$

Part (b)

Step 1. Expand the brack­ets term by term (first, outer, inner, last).

$$(4 + i)(2 - 3i) = 8 - 12i + 2i - 3i^2$$

Step 2. Replace $i^2$ by $-1$, so $-3i^2 = +3$.

$$= 8 - 12i + 2i + 3$$

Step 3. Col­lect real and imag­i­nary parts.

$$= 11 - 10i$$

Part (c)

Step 1. Divide the expo­nent by 4: $27 = 4 \times 6 + 3$, so the remain­der is 3.

$$i^{27} = i^{24} \cdot i^{3} = (i^4)^6 \cdot i^3 = 1 \cdot i^3$$

Step 2. Use $i^3 = i^2 \cdot i = -i$.

$$i^{27} = -i = 0 - i$$

Part (d)

Step 1. A neg­a­tive power is a rec­i­p­ro­cal.

$$\displaystyle i^{-18} = \frac{1}{i^{18}}$$

Step 2. $18 = 4 \times 4 + 2$, so $i^{18} = i^2 = -1$.

$$\displaystyle i^{-18} = \frac{1}{-1} = -1 = -1 + 0i$$

Part (e)

Step 1. First square $2 - i$.

$$(2 - i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i$$

Step 2. Mul­ti­ply the result by $2 - i$ once more.

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

Check­ing the answer#

For (b), the mod­u­lus of a prod­uct is the prod­uct of mod­uli: $\lvert 4 + i \rvert \lvert 2 - 3i \rvert = \sqrt{17}\sqrt{13} = \sqrt{221}$, and $\lvert 11 - 10i \rvert = \sqrt{121 + 100} = \sqrt{221}$. For (e), $\lvert 2 - i \rvert^3 = (\sqrt5)^3 = 5\sqrt5 = \sqrt{125}$, and $\lvert 2 - 11i \rvert = \sqrt{4 + 121} = \sqrt{125}$. Both agree.

Answer#

(a) $3 + 4i$; (b) $11 - 10i$; (c) $-i$; (d) $-1$; (e) $2 - 11i$.

Com­mon mis­take to avoid#

For­get­ting to turn $i^2$ into $-1$ (so $-3i^2$ becomes $+3$, not $-3$) is the most com­mon slip in part (b).

Ques­tion 2: Find­ing real x and y by com­par­ing parts#

The prob­lem#

Find real $x, y$: (a) $(x + y) + i(x - y) = 6 + 2i$; (b) $(3 + 2i)x - (1 - i)y = 5 + 5i$.

Under­stand­ing the prob­lem#

Here $x$ and $y$ are real num­bers. Two com­plex num­bers are equal exactly when their real parts are equal and their imag­i­nary parts are equal. So one com­plex equa­tion gives two ordi­nary equa­tions.

The idea#

Write the left-hand side in the form (real part) $+ i\,$(imag­i­nary part), then equate real parts and imag­i­nary parts, and solve the pair of lin­ear equa­tions.

Step-by-step solu­tion#

Part (a)

Step 1. The left side is already in the form $a + ib$ with $a = x + y$ and $b = x - y$. Com­pare with $6 + 2i$.

$$x + y = 6, \qquad x - y = 2$$

Step 2. Add the two equa­tions to remove $y$.

$$2x = 8 \quad\Rightarrow\quad x = 4$$

Step 3. Sub­sti­tute back.

$$4 + y = 6 \quad\Rightarrow\quad y = 2$$

Part (b)

Step 1. Expand the left side, remem­ber­ing $x$ and $y$ are real.

$$(3 + 2i)x - (1 - i)y = 3x + 2ix - y + iy = (3x - y) + i(2x + y)$$

Step 2. Equate real and imag­i­nary parts with $5 + 5i$.

$$3x - y = 5, \qquad 2x + y = 5$$

Step 3. Add the two equa­tions.

$$5x = 10 \quad\Rightarrow\quad x = 2$$

Step 4. Sub­sti­tute into $2x + y = 5$.

$$4 + y = 5 \quad\Rightarrow\quad y = 1$$

Check­ing the answer#

(a) $(4 + 2) + i(4 - 2) = 6 + 2i$. ✓
(b) $(3 + 2i)(2) - (1 - i)(1) = 6 + 4i - 1 + i = 5 + 5i$. ✓

Answer#

(a) $x = 4$, $y = 2$; (b) $x = 2$, $y = 1$.

Com­mon mis­take to avoid#

In (b), the minus sign in front of $(1 - i)y$ applies to both terms: $-(1 - i)y = -y + iy$.

Ques­tion 3: Con­ju­gate, mod­u­lus and inverse#

The prob­lem#

Find the con­ju­gate, mod­u­lus and mul­ti­plica­tive inverse of $-5 + 12i$.

Under­stand­ing the prob­lem#

For $z = a + ib$: the con­ju­gate $\bar z$ flips the sign of the imag­i­nary part; the mod­u­lus $\lvert z \rvert$ is its dis­tance from $0$; the mul­ti­plica­tive inverse $\displaystyle \tfrac{1}{z}$ is the num­ber that mul­ti­plies $z$ to give $1$. Here $a = -5$, $b = 12$.

The idea#

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

Step-by-step solu­tion#

Step 1. Con­ju­gate: change the sign of the imag­i­nary part only.

$$\bar z = -5 - 12i$$

Step 2. Mod­u­lus.

$$\lvert z \rvert = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$

Step 3. Inverse: divide the con­ju­gate by $\lvert z \rvert^2 = 169$.

$$\displaystyle \frac{1}{z} = \frac{\bar z}{\lvert z \rvert^2} = \frac{-5 - 12i}{169} = -\frac{5}{169} - \frac{12}{169}i$$

Check­ing the answer#

Mul­ti­ply $z$ by the inverse: $(-5 + 12i)(-5 - 12i) = 25 - 144i^2 = 25 + 144 = 169$, and $\displaystyle \tfrac{169}{169} = 1$. ✓

Answer#

Con­ju­gate $-5 - 12i$; mod­u­lus $13$; inverse $\displaystyle \frac{-5 - 12i}{169}$.

Ques­tion 4: Divi­sion of com­plex num­bers#

The prob­lem#

Sim­plify (a) $\displaystyle \frac{3 - 4i}{2 + i}$ and (b) $\displaystyle \frac{(1 + i)^2}{3 - i}$.

Under­stand­ing the prob­lem#

A com­plex num­ber is not in stan­dard form while $i$ sits in the denom­i­na­tor. You must rewrite each frac­tion as $a + ib$.

The idea#

Mul­ti­ply the numer­a­tor and denom­i­na­tor by the con­ju­gate of the denom­i­na­tor. The denom­i­na­tor becomes $c^2 + d^2$, a real num­ber.

Step-by-step solu­tion#

Part (a)

Step 1. The denom­i­na­tor is $2 + i$, so mul­ti­ply top and bot­tom by $2 - i$.

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

Step 2. Denom­i­na­tor: $(2 + i)(2 - i) = 4 - i^2 = 4 + 1 = 5$.

Step 3. Numer­a­tor.

$$(3 - 4i)(2 - i) = 6 - 3i - 8i + 4i^2 = 6 - 11i - 4 = 2 - 11i$$

Step 4. Put them together.

$$\displaystyle \frac{3 - 4i}{2 + i} = \frac{2 - 11i}{5} = \frac{2}{5} - \frac{11}{5}i$$

Part (b)

Step 1. Sim­plify the numer­a­tor first.

$$(1 + i)^2 = 1 + 2i + i^2 = 2i$$

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

$$\displaystyle \frac{2i}{3 - i} = \frac{2i(3 + i)}{(3 - i)(3 + i)} = \frac{6i + 2i^2}{9 + 1}$$

Step 3. Sim­plify.

$$\displaystyle = \frac{-2 + 6i}{10} = \frac{-1 + 3i}{5} = -\frac{1}{5} + \frac{3}{5}i$$

Check­ing the answer#

(a) Mul­ti­ply back: $\displaystyle \tfrac{2 - 11i}{5}(2 + i) = \tfrac{4 + 2i - 22i - 11i^2}{5} = \tfrac{15 - 20i}{5} = 3 - 4i$. ✓
(b) $\displaystyle \tfrac{-1 + 3i}{5}(3 - i) = \tfrac{-3 + i + 9i - 3i^2}{5} = \tfrac{10i}{5} = 2i$. ✓

Answer#

(a) $\displaystyle \frac{2 - 11i}{5}$; (b) $\displaystyle \frac{-1 + 3i}{5}$.

Ques­tion 5: Square roots of neg­a­tive num­bers#

The prob­lem#

Eval­u­ate $\sqrt{-16} + 3\sqrt{-25} - \sqrt{-49}$.

Under­stand­ing the prob­lem#

Each square root has a neg­a­tive num­ber inside. For a pos­i­tive real $a$, $\sqrt{-a} = i\sqrt{a}$.

The idea#

Turn each root into a real mul­ti­ple of $i$, then add like terms.

Step-by-step solu­tion#

Step 1. Rewrite each root.

$$\sqrt{-16} = 4i, \qquad \sqrt{-25} = 5i, \qquad \sqrt{-49} = 7i$$

Step 2. Sub­sti­tute and com­bine.

$$4i + 3(5i) - 7i = 4i + 15i - 7i = 12i$$

Check­ing the answer#

Every term is purely imag­i­nary, so the answer must be purely imag­i­nary; the coef­fi­cients $4 + 15 - 7 = 12$ agree.

Answer#

$12i$

Ques­tion 6: Sum of four con­sec­u­tive pow­ers of i#

The prob­lem#

Show that $i^{n} + i^{n+1} + i^{n+2} + i^{n+3} = 0$ for every inte­ger $n$.

Under­stand­ing the prob­lem#

This is a proof: you must show the sum is zero no mat­ter which inte­ger $n$ is, pos­i­tive, neg­a­tive or zero.

The idea#

Every term con­tains the com­mon fac­tor $i^n$. Take it out; what remains is $1 + i + i^2 + i^3$, which you can com­pute directly.

Step-by-step solu­tion#

Step 1. Fac­tor out $i^n$ (valid because $i^{n+k} = i^n \cdot i^k$ for every inte­ger $n$).

$$i^{n} + i^{n+1} + i^{n+2} + i^{n+3} = i^n\left(1 + i + i^2 + i^3\right)$$

Step 2. Use $i^2 = -1$ and $i^3 = -i$.

$$1 + i + i^2 + i^3 = 1 + i - 1 - i = 0$$

Step 3. There­fore the whole sum is $i^n \times 0 = 0$, for every inte­ger $n$.

Check­ing the answer#

Try $n = 1$: $i + (-1) + (-i) + 1 = 0$. ✓

Answer#

$i^{n} + i^{n+1} + i^{n+2} + i^{n+3} = i^n(1 + i - 1 - i) = 0$ for every inte­ger $n$.

Ques­tion 7: A large power of a quo­tient#

The prob­lem#

Find $\displaystyle \left(\frac{1 + i}{1 - i}\right)^{50}$.

Under­stand­ing the prob­lem#

Expand­ing a 50th power directly is impos­si­ble by hand. First sim­plify the quo­tient inside the bracket; it turns out to be a power of $i$.

The idea#

Divide using the con­ju­gate, then use the cycle of pow­ers of $i$.

Step-by-step solu­tion#

Step 1. Mul­ti­ply top and bot­tom by $1 + i$.

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

Step 2. So the expres­sion is $i^{50}$. Divide 50 by 4: $50 = 4 \times 12 + 2$.

$$i^{50} = (i^4)^{12}\, i^2 = 1 \cdot (-1) = -1$$

Check­ing the answer#

$\displaystyle \left\lvert \tfrac{1 + i}{1 - i} \right\rvert = \tfrac{\sqrt2}{\sqrt2} = 1$, so the 50th power must also have mod­u­lus 1; $\lvert -1 \rvert = 1$. ✓

Answer#

$-1$

Ques­tion 8: Mod­u­lus of a prod­uct and a real part#

The prob­lem#

If $z_1 = 2 - i$ and $z_2 = 1 + 3i$, ver­ify $\lvert z_1 z_2 \rvert = \lvert z_1 \rvert \lvert z_2 \rvert$ and find $\displaystyle \operatorname{Re}\left(\frac{z_1}{z_2}\right)$.

Under­stand­ing the prob­lem#

There are two tasks. First, cal­cu­late both sides of $\lvert z_1 z_2 \rvert = \lvert z_1 \rvert \lvert z_2 \rvert$ sep­a­rately and see they are equal. Sec­ond, write $\displaystyle \tfrac{z_1}{z_2}$ as $a + ib$ and state $a$.

The idea#

Mul­ti­ply out $z_1 z_2$ and use $\lvert a + ib \rvert = \sqrt{a^2 + b^2}$. For the quo­tient, mul­ti­ply top and bot­tom by $\bar z_2 = 1 - 3i$.

Step-by-step solu­tion#

Part (a): ver­i­fy­ing the mod­u­lus prop­erty

Step 1. Mul­ti­ply.

$$z_1 z_2 = (2 - i)(1 + 3i) = 2 + 6i - i - 3i^2 = 2 + 5i + 3 = 5 + 5i$$

Step 2. Left side.

$$\lvert z_1 z_2 \rvert = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt2$$

Step 3. Right side.

$$\lvert z_1 \rvert \lvert z_2 \rvert = \sqrt{4 + 1}\,\sqrt{1 + 9} = \sqrt5\sqrt{10} = \sqrt{50} = 5\sqrt2$$

Both sides equal $5\sqrt2$, so the prop­erty holds.

Part (b): the real part of the quo­tient

Step 1. Mul­ti­ply numer­a­tor and denom­i­na­tor by $1 - 3i$.

$$\displaystyle \frac{z_1}{z_2} = \frac{(2 - i)(1 - 3i)}{(1 + 3i)(1 - 3i)}$$

Step 2. Denom­i­na­tor: $1 + 9 = 10$. Numer­a­tor:

$$(2 - i)(1 - 3i) = 2 - 6i - i + 3i^2 = 2 - 7i - 3 = -1 - 7i$$

Step 3. So

$$\displaystyle \frac{z_1}{z_2} = \frac{-1 - 7i}{10} = -\frac{1}{10} - \frac{7}{10}i$$

and the real part is $\displaystyle -\tfrac{1}{10}$.

Check­ing the answer#

$\displaystyle \left\lvert \tfrac{-1 - 7i}{10} \right\rvert = \tfrac{\sqrt{50}}{10} = \tfrac{\sqrt2}{2}$, and $\displaystyle \tfrac{\lvert z_1 \rvert}{\lvert z_2 \rvert} = \tfrac{\sqrt5}{\sqrt{10}} = \tfrac{1}{\sqrt2} = \tfrac{\sqrt2}{2}$. ✓

Answer#

$\lvert z_1 z_2 \rvert = 5\sqrt2 = \lvert z_1 \rvert \lvert z_2 \rvert$; $\displaystyle \operatorname{Re}\left(\tfrac{z_1}{z_2}\right) = -\tfrac{1}{10}$.

Ques­tion 9: z plus and minus its con­ju­gate#

The prob­lem#

Prove $z + \bar z$ is real and $z - \bar z$ is purely imag­i­nary.

Under­stand­ing the prob­lem#

"Real" means the imag­i­nary part is $0$. "Purely imag­i­nary" means the real part is $0$. You must show this holds for every com­plex num­ber $z$.

The idea#

Write a gen­eral $z = a + ib$ with $a, b$ real, so $\bar z = a - ib$, and just add and sub­tract.

Step-by-step solu­tion#

Step 1. Let $z = a + ib$, where $a, b$ are real. Then $\bar z = a - ib$.

Step 2. Add.

$$z + \bar z = (a + ib) + (a - ib) = 2a$$

$2a$ is a real num­ber (its imag­i­nary part is $0$), so $z + \bar z$ is real.

Step 3. Sub­tract.

$$z - \bar z = (a + ib) - (a - ib) = 2ib$$

$2ib = 0 + i(2b)$ has real part $0$, so $z - \bar z$ is purely imag­i­nary.

Check­ing the answer#

Take $z = 3 + 4i$: $z + \bar z = 6$ (real) and $z - \bar z = 8i$ (purely imag­i­nary). ✓

Answer#

$z + \bar z = 2\operatorname{Re} z$ is real and $z - \bar z = 2i\operatorname{Im} z$ is purely imag­i­nary.

Ques­tion 10: The least power that gives 1#

The prob­lem#

Find the least pos­i­tive inte­ger $m$ with $\displaystyle \left(\frac{1 + i}{1 - i}\right)^m = 1$.

Under­stand­ing the prob­lem#

You need the small­est $m = 1, 2, 3, \dots$ that makes the power equal to exactly $1$.

The idea#

From Ques­tion 7, $\displaystyle \frac{1 + i}{1 - i} = i$. So you need the least pos­i­tive $m$ with $i^m = 1$.

Step-by-step solu­tion#

Step 1. Sim­plify the base (as in Ques­tion 7).

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

Step 2. List the pow­ers of $i$ in order.

$$i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1$$

Step 3. The first power equal to $1$ is the fourth, so $m = 4$.

Check­ing the answer#

$m = 1, 2, 3$ give $i, -1, -i$, none of which is $1$, so no smaller $m$ works.

Answer#

$m = 4$