Argand Plane, Polar Form and Quadratics, Worked Solutions

Ques­tion 1: Plot­ting a num­ber, its con­ju­gate and its neg­a­tive#

The prob­lem#

Plot $3 + 2i$, its con­ju­gate and its neg­a­tive on the Argand plane.

Under­stand­ing the prob­lem#

You are given one com­plex num­ber, $z = 3 + 2i$. You must find two related num­bers, the con­ju­gate $\bar z$ and the neg­a­tive $-z$, and mark all three as points on the Argand plane. Remem­ber that on the Argand plane the num­ber $a + ib$ is the point $(a, b)$: the real part goes along the hor­i­zon­tal axis and the imag­i­nary part along the ver­ti­cal axis.

The idea#

The les­son tells you that $\bar z$ is the reflec­tion of $z$ in the real axis (only the sign of the imag­i­nary part changes), and $-z$ is the reflec­tion of $z$ through the ori­gin (both signs change). So we write each num­ber in the form $a + ib$ and read off its point $(a, b)$.

Step-by-step solu­tion#

Step 1. Write $z$ as a point. The real part is $3$ and the imag­i­nary part is $2$.

$$z = 3 + 2i \;\longrightarrow\; (3, 2)$$

This point lies in the first quad­rant.

Step 2. Find the con­ju­gate by chang­ing the sign of the imag­i­nary part only.

$$\bar z = 3 - 2i \;\longrightarrow\; (3, -2)$$

This is the mir­ror image of $(3, 2)$ in the real axis, in the fourth quad­rant.

Step 3. Find the neg­a­tive by chang­ing the sign of both parts.

$$-z = -3 - 2i \;\longrightarrow\; (-3, -2)$$

This is the point dia­met­ri­cally oppo­site $(3, 2)$ through the ori­gin, in the third quad­rant.

Step 4. Draw the pic­ture. Draw the real axis (hor­i­zon­tal) and imag­i­nary axis (ver­ti­cal). Mark $(3, 2)$ and label it $z$; mark $(3, -2)$ directly below it and label it $\bar z$; mark $(-3, -2)$ and label it $-z$. If you join $z$ to $-z$, the seg­ment passes through the ori­gin $O$, and $O$ is its mid­point.

Check­ing the answer#

All three points are the same dis­tance from the ori­gin, $\sqrt{3^2 + 2^2} = \sqrt{13}$, as they must be, because reflec­tions do not change dis­tance from $O$.

Answer#

$z = (3, 2)$, $\bar z = (3, -2)$, $-z = (-3, -2)$.

Com­mon mis­take to avoid#

Do not con­fuse $\bar z$ with $-z$. The con­ju­gate changes only the sign of $i$; the neg­a­tive changes both signs.

Ques­tion 2: Dis­tance between two com­plex num­bers#

The prob­lem#

Find the dis­tance between $1 + 2i$ and $4 - 2i$.

Under­stand­ing the prob­lem#

The two num­bers are the points $(1, 2)$ and $(4, -2)$ on the Argand plane. You are asked for the straight-line dis­tance between these points.

The idea#

The les­son states that $\lvert z_1 - z_2 \rvert$ is the dis­tance between the points $z_1$ and $z_2$. So we sub­tract the num­bers and find the mod­u­lus of the result, using $\lvert a + ib \rvert = \sqrt{a^2 + b^2}$.

Step-by-step solu­tion#

Step 1. Sub­tract the two num­bers, tak­ing real parts and imag­i­nary parts sep­a­rately.

$$\begin{aligned} z_1 - z_2 &= (1 + 2i) - (4 - 2i) \\ &= (1 - 4) + (2 - (-2))i \\ &= -3 + 4i \end{aligned}$$

Step 2. Take the mod­u­lus of the dif­fer­ence.

$$\lvert -3 + 4i \rvert = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Check­ing the answer#

Using the dis­tance for­mula on $(1, 2)$ and $(4, -2)$: $\sqrt{(4 - 1)^2 + (-2 - 2)^2} = \sqrt{9 + 16} = 5$. The two meth­ods agree.

Answer#

The dis­tance is $5$.

Com­mon mis­take to avoid#

When sub­tract­ing $-2i$, the sign changes: $2i - (-2i) = 4i$, not $0$.

Ques­tion 3: Mod­u­lus and prin­ci­pal argu­ment#

The prob­lem#

Find the mod­u­lus and prin­ci­pal argu­ment of: (a) $1 - i$; (b) $-\sqrt{3} + i$; (c) $-2i$; (d) $-\sqrt{3} - i$.

Under­stand­ing the prob­lem#

For each num­ber $x + iy$ you need two things: the mod­u­lus $r = \sqrt{x^2 + y^2}$ (its dis­tance from the ori­gin), and the prin­ci­pal argu­ment $\theta$, the angle it makes with the pos­i­tive real axis, cho­sen in the inter­val $(-\pi, \pi]$. The con­di­tion "prin­ci­pal" mat­ters: angles below the real axis must be given as neg­a­tive angles, not as angles like $\displaystyle \tfrac{7\pi}{4}$.

The idea#

Use the lesson's method: find the ref­er­ence angle $\displaystyle \alpha = \tan^{-1}\left\lvert \tfrac{y}{x} \right\rvert$, which is always between $0$ and $\displaystyle \tfrac{\pi}{2}$, then place it by quad­rant:

• Quad­rant I: $\theta = \alpha$
• Quad­rant II: $\theta = \pi - \alpha$
• Quad­rant III: $\theta = -(\pi - \alpha)$
• Quad­rant IV: $\theta = -\alpha$

Points on an axis are han­dled directly by look­ing at where they lie.

Step-by-step solu­tion#

Part (a): $1 - i$

Step 1. Iden­tify $x = 1$, $y = -1$. Find the mod­u­lus.

$$r = \sqrt{1^2 + (-1)^2} = \sqrt{2}$$

Step 2. Find the ref­er­ence angle.

$$\displaystyle \alpha = \tan^{-1}\left\lvert \frac{-1}{1} \right\rvert = \tan^{-1} 1 = \frac{\pi}{4}$$

Step 3. Place it by quad­rant. $x > 0$ and $y < 0$, so the point $(1, -1)$ is in quad­rant IV, and $\theta = -\alpha$.

$$\displaystyle \theta = -\frac{\pi}{4}$$

Part (b): $-\sqrt{3} + i$

Step 1. Here $x = -\sqrt{3}$, $y = 1$.

$$r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$$

Step 2. Ref­er­ence angle:

$$\displaystyle \alpha = \tan^{-1}\left\lvert \frac{1}{-\sqrt{3}} \right\rvert = \tan^{-1}\frac{1}{\sqrt{3}} = \frac{\pi}{6}$$

Step 3. $x < 0$, $y > 0$: quad­rant II, so $\theta = \pi - \alpha$.

$$\displaystyle \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$

Part (c): $-2i$

Step 1. Here $x = 0$, $y = -2$, so the point is $(0, -2)$.

$$r = \sqrt{0^2 + (-2)^2} = 2$$

Step 2. We can­not use $\displaystyle \tfrac{y}{x}$ because $x = 0$. Instead, look at the posi­tion: $(0, -2)$ lies on the neg­a­tive imag­i­nary axis, which points straight down, a quar­ter turn clock­wise from the pos­i­tive real axis.

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

Part (d): $-\sqrt{3} - i$

Step 1. Here $x = -\sqrt{3}$, $y = -1$.

$$r = \sqrt{3 + 1} = 2$$

Step 2. Ref­er­ence angle, exactly as in part (b):

$$\displaystyle \alpha = \tan^{-1}\frac{1}{\sqrt{3}} = \frac{\pi}{6}$$

Step 3. $x < 0$, $y < 0$: quad­rant III, so $\theta = -(\pi - \alpha)$.

$$\displaystyle \theta = -\left(\pi - \frac{\pi}{6}\right) = -\frac{5\pi}{6}$$

Check­ing the answer#

Rebuild each num­ber from $r(\cos\theta + i\sin\theta)$. For (d): $\displaystyle 2\left(\cos\left(-\tfrac{5\pi}{6}\right) + i\sin\left(-\tfrac{5\pi}{6}\right)\right) = 2\left(-\tfrac{\sqrt{3}}{2} - \tfrac{1}{2}i\right) = -\sqrt{3} - i$. Cor­rect. Also every $\theta$ lies in $(-\pi, \pi]$, as a prin­ci­pal argu­ment must.

Answer#

(a) $\sqrt{2}$, $\displaystyle -\tfrac{\pi}{4}$; (b) $2$, $\displaystyle \tfrac{5\pi}{6}$; (c) $2$, $\displaystyle -\tfrac{\pi}{2}$; (d) $2$, $\displaystyle -\tfrac{5\pi}{6}$.

Com­mon mis­take to avoid#

A cal­cu­la­tor's $\displaystyle \tan^{-1}\tfrac{y}{x}$ gives the same value for $-\sqrt{3} + i$ and $\sqrt{3} - i$. Always decide the quad­rant from the signs of $x$ and $y$ first.

Ques­tion 4: Writ­ing num­bers in polar form#

The prob­lem#

Write in polar form: (a) $\sqrt{3} + i$; (b) $-1 - i$; (c) $6$; (d) $\displaystyle \frac{1 + 3i}{1 - 2i}$.

Under­stand­ing the prob­lem#

Polar form means writ­ing the num­ber as $r(\cos\theta + i\sin\theta)$, where $r$ is the mod­u­lus and $\theta$ the prin­ci­pal argu­ment. In part (d) the num­ber is a quo­tient, so it is not yet in the form $x + iy$; you must sim­plify it first.

The idea#

For each num­ber: get it into the form $x + iy$, find $r$ and $\theta$ exactly as in Ques­tion 3, then sub­sti­tute into $r(\cos\theta + i\sin\theta)$. To sim­plify a quo­tient, mul­ti­ply numer­a­tor and denom­i­na­tor by the con­ju­gate of the denom­i­na­tor, as in Exam­ple 4 of the les­son.

Step-by-step solu­tion#

Part (a): $\sqrt{3} + i$

Step 1. Mod­u­lus: $r = \sqrt{3 + 1} = 2$.

Step 2. Ref­er­ence angle $\displaystyle \alpha = \tan^{-1}\tfrac{1}{\sqrt{3}} = \tfrac{\pi}{6}$. Both parts are pos­i­tive, so the point is in quad­rant I and $\displaystyle \theta = \tfrac{\pi}{6}$.

Step 3. Sub­sti­tute.

$$\displaystyle \sqrt{3} + i = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$$

Part (b): $-1 - i$

Step 1. Mod­u­lus: $r = \sqrt{1 + 1} = \sqrt{2}$.

Step 2. Ref­er­ence angle $\displaystyle \alpha = \tan^{-1} 1 = \tfrac{\pi}{4}$. Both parts are neg­a­tive, so quad­rant III, and

$$\displaystyle \theta = -\left(\pi - \frac{\pi}{4}\right) = -\frac{3\pi}{4}$$

Step 3. Sub­sti­tute.

$$\displaystyle -1 - i = \sqrt{2}\left(\cos\left(-\frac{3\pi}{4}\right) + i\sin\left(-\frac{3\pi}{4}\right)\right)$$

Part (c): $6$

Step 1. $6 = 6 + 0i$ is the point $(6, 0)$ on the pos­i­tive real axis. Its dis­tance from the ori­gin is $r = 6$.

Step 2. A point on the pos­i­tive real axis makes no angle with it, so $\theta = 0$.

$$6 = 6(\cos 0 + i\sin 0)$$

Part (d): $\displaystyle \frac{1 + 3i}{1 - 2i}$

Step 1. Remove $i$ from the denom­i­na­tor by mul­ti­ply­ing top and bot­tom by the con­ju­gate of $1 - 2i$, which is $1 + 2i$.

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

Step 2. Expand the numer­a­tor, remem­ber­ing $i^2 = -1$.

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

Step 3. Expand the denom­i­na­tor using $(a - ib)(a + ib) = a^2 + b^2$.

$$(1 - 2i)(1 + 2i) = 1^2 + 2^2 = 5$$

Step 4. Divide.

$$\displaystyle \frac{-5 + 5i}{5} = -1 + i$$

Step 5. Now find $r$ and $\theta$ for $-1 + i$: $r = \sqrt{1 + 1} = \sqrt{2}$, ref­er­ence angle $\displaystyle \tfrac{\pi}{4}$, and $x < 0$, $y > 0$ puts it in quad­rant II, so $\displaystyle \theta = \pi - \tfrac{\pi}{4} = \tfrac{3\pi}{4}$.

$$\displaystyle \frac{1 + 3i}{1 - 2i} = \sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)$$

Check­ing the answer#

For (d), mul­ti­ply back: $(-1 + i)(1 - 2i) = -1 + 2i + i - 2i^2 = -1 + 3i + 2 = 1 + 3i$. So $-1 + i$ really is the quo­tient. For (b), $\displaystyle \sqrt{2}\left(-\tfrac{1}{\sqrt{2}} - \tfrac{1}{\sqrt{2}}i\right) = -1 - i$.

Answer#

(a) $\displaystyle 2\left(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}\right)$; (b) $\displaystyle \sqrt{2}\left(\cos\left(-\tfrac{3\pi}{4}\right) + i\sin\left(-\tfrac{3\pi}{4}\right)\right)$; (c) $6(\cos 0 + i\sin 0)$; (d) $\displaystyle \sqrt{2}\left(\cos\tfrac{3\pi}{4} + i\sin\tfrac{3\pi}{4}\right)$.

Com­mon mis­take to avoid#

In (b), do not write $\displaystyle \theta = \tfrac{5\pi}{4}$. That angle is cor­rect as a direc­tion but is out­side $(-\pi, \pi]$, so it is not the prin­ci­pal argu­ment.

Ques­tion 5: Qua­drat­ics with neg­a­tive dis­crim­i­nant#

The prob­lem#

Solve: (a) $x^2 + 9 = 0$; (b) $x^2 + 2x + 5 = 0$; (c) $2x^2 - 3x + 4 = 0$; (d) $\displaystyle x^2 + \frac{x}{\sqrt{3}} + 1 = 0$.

Under­stand­ing the prob­lem#

Each is a qua­dratic equa­tion with real coef­fi­cients. You must find both roots. None of these has real roots, so the roots will be com­plex num­bers, and they will come as a con­ju­gate pair.

The idea#

Write each equa­tion as $ax^2 + bx + c = 0$, com­pute the dis­crim­i­nant $D = b^2 - 4ac$, and, since $D < 0$, use the lesson's for­mula

$$\displaystyle x = \frac{-b \pm i\sqrt{4ac - b^2}}{2a}.$$

In (d), clear­ing the frac­tion first makes the arith­metic eas­ier.

Step-by-step solu­tion#

Part (a): $x^2 + 9 = 0$

Step 1. There is no $x$ term, so it is quick­est to iso­late $x^2$.

$$x^2 = -9$$

Step 2. The num­bers whose square is $-9$ are $3i$ and $-3i$, because $(3i)^2 = 9i^2 = -9$.

$$x = \pm 3i$$

Part (b): $x^2 + 2x + 5 = 0$

Step 1. Read off $a = 1$, $b = 2$, $c = 5$ and find the dis­crim­i­nant.

$$D = 2^2 - 4(1)(5) = 4 - 20 = -16$$

Step 2. $D < 0$, so $4ac - b^2 = 16$ and $\sqrt{16} = 4$. Apply the for­mula.

$$\displaystyle x = \frac{-2 \pm 4i}{2(1)} = -1 \pm 2i$$

Part (c): $2x^2 - 3x + 4 = 0$

Step 1. $a = 2$, $b = -3$, $c = 4$.

$$D = (-3)^2 - 4(2)(4) = 9 - 32 = -23$$

Step 2. $4ac - b^2 = 23$, which is not a per­fect square, so leave it as $\sqrt{23}$. Note $-b = 3$.

$$\displaystyle x = \frac{3 \pm \sqrt{23}\,i}{2(2)} = \frac{3 \pm \sqrt{23}\,i}{4}$$

Part (d): $\displaystyle x^2 + \frac{x}{\sqrt{3}} + 1 = 0$

Step 1. Mul­ti­ply every term by $\sqrt{3}$ to clear the frac­tion. This does not change the roots.

$$\sqrt{3}x^2 + x + \sqrt{3} = 0$$

Step 2. Now $a = \sqrt{3}$, $b = 1$, $c = \sqrt{3}$.

$$D = 1^2 - 4(\sqrt{3})(\sqrt{3}) = 1 - 12 = -11$$

Step 3. $4ac - b^2 = 11$. Apply the for­mula.

$$\displaystyle x = \frac{-1 \pm \sqrt{11}\,i}{2\sqrt{3}}$$

Check­ing the answer#

Check (b) by sub­sti­tut­ing $x = -1 + 2i$: $x^2 = 1 - 4i + 4i^2 = -3 - 4i$, and $2x = -2 + 4i$, so $x^2 + 2x + 5 = -3 - 4i - 2 + 4i + 5 = 0$. You can also check any pair with sum and prod­uct of roots: in (c), sum $\displaystyle = \tfrac{6}{4} = \tfrac{3}{2} = -\tfrac{b}{a}$ and prod­uct $\displaystyle = \tfrac{9 + 23}{16} = 2 = \tfrac{c}{a}$.

Answer#

(a) $\pm 3i$; (b) $-1 \pm 2i$; (c) $\displaystyle \frac{3 \pm \sqrt{23}\,i}{4}$; (d) $\displaystyle \frac{-1 \pm \sqrt{11}\,i}{2\sqrt{3}}$.

Com­mon mis­take to avoid#

In (c), $b = -3$, so $-b = +3$, and $b^2 = +9$. Sign slips on $b$ are the most com­mon error with this for­mula.

Ques­tion 6: Show­ing roots are con­ju­gates#

The prob­lem#

Show that the roots of $x^2 - 6x + 10 = 0$ are con­ju­gates and find their prod­uct.

Under­stand­ing the prob­lem#

You must find the two roots, show that each is the con­ju­gate of the other (same real part, oppo­site imag­i­nary parts), and then mul­ti­ply them.

The idea#

Solve with the for­mula for a neg­a­tive dis­crim­i­nant, then com­pare the roots. For the prod­uct, mul­ti­ply directly, and check with the fact that the prod­uct of the roots of $ax^2 + bx + c = 0$ is $\displaystyle \tfrac{c}{a}$.

Step-by-step solu­tion#

Step 1. Here $a = 1$, $b = -6$, $c = 10$.

$$D = (-6)^2 - 4(1)(10) = 36 - 40 = -4$$

Step 2. $4ac - b^2 = 4$, $\sqrt{4} = 2$, and $-b = 6$.

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

Step 3. Com­pare the roots. The roots are $3 + i$ and $3 - i$. They have the same real part $3$ and imag­i­nary parts $1$ and $-1$, so each is the con­ju­gate of the other. This is what the les­son pre­dicts for any real qua­dratic with $D < 0$.

Step 4. Mul­ti­ply, using $(a + ib)(a - ib) = a^2 + b^2$.

$$(3 + i)(3 - i) = 3^2 + 1^2 = 10$$

Check­ing the answer#

The prod­uct of the roots should equal $\displaystyle \tfrac{c}{a} = \tfrac{10}{1} = 10$. It does.

Answer#

The roots are $3 + i$ and $3 - i$, a con­ju­gate pair, and their prod­uct is $10$.

Ques­tion 7: Build­ing a qua­dratic from a com­plex root#

The prob­lem#

Find a qua­dratic with real coef­fi­cients hav­ing $3 - 2i$ as a root.

Under­stand­ing the prob­lem#

You are given only one root. The con­di­tion "real coef­fi­cients" is the key: it tells you some­thing about the other root.

The idea#

The roots of a real qua­dratic with $D < 0$ come as a con­ju­gate pair, so the sec­ond root must be $3 + 2i$. A qua­dratic with roots $\alpha$ and $\beta$ is

$$x^2 - (\alpha + \beta)x + \alpha\beta = 0.$$

Step-by-step solu­tion#

Step 1. Write down the other root, the con­ju­gate of $3 - 2i$.

$$\beta = 3 + 2i$$

Step 2. Find the sum of the roots; the imag­i­nary parts can­cel.

$$\alpha + \beta = (3 - 2i) + (3 + 2i) = 6$$

Step 3. Find the prod­uct.

$$\alpha\beta = (3 - 2i)(3 + 2i) = 9 + 4 = 13$$

Step 4. Sub­sti­tute into $x^2 - (\text{sum})x + (\text{product}) = 0$.

$$x^2 - 6x + 13 = 0$$

Check­ing the answer#

Solve it: $D = 36 - 52 = -16$, so $\displaystyle x = \tfrac{6 \pm 4i}{2} = 3 \pm 2i$. One root is $3 - 2i$, as required, and the coef­fi­cients $1, -6, 13$ are real.

Answer#

$x^2 - 6x + 13 = 0$.

Ques­tion 8: Con­ju­gate and rec­i­p­ro­cal in polar form#

The prob­lem#

If $z = r(\cos\theta + i\sin\theta)$, write $\bar z$ and $\displaystyle \frac{1}{z}$ in polar form.

Under­stand­ing the prob­lem#

You are given $z$ in polar form with mod­u­lus $r > 0$ and argu­ment $\theta$. You must express its con­ju­gate and its rec­i­p­ro­cal also in the form (mod­u­lus)$(\cos(\text{angle}) + i\sin(\text{angle}))$.

The idea#

Two facts do the work: $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$. For the rec­i­p­ro­cal, mul­ti­ply numer­a­tor and denom­i­na­tor by $\bar z$, since $z\bar z = \lvert z \rvert^2 = r^2$.

Step-by-step solu­tion#

Step 1. Write $z$ in the form $x + iy$.

$$z = r\cos\theta + i\,r\sin\theta$$

Step 2. The con­ju­gate changes the sign of the imag­i­nary part.

$$\bar z = r\cos\theta - i\,r\sin\theta = r(\cos\theta - i\sin\theta)$$

Step 3. This is not yet polar form, because polar form needs a plus sign. Use $\cos\theta = \cos(-\theta)$ and $-\sin\theta = \sin(-\theta)$.

$$\bar z = r\left(\cos(-\theta) + i\sin(-\theta)\right)$$

So $\bar z$ has the same mod­u­lus $r$ and argu­ment $-\theta$: it is the reflec­tion of $z$ in the real axis.

Step 4. For the rec­i­p­ro­cal, mul­ti­ply top and bot­tom by $\bar z$.

$$\displaystyle \frac{1}{z} = \frac{\bar z}{z\bar z} = \frac{\bar z}{r^2}$$

Step 5. Sub­sti­tute the polar form of $\bar z$ from Step 3 and sim­plify $\displaystyle \tfrac{r}{r^2} = \tfrac{1}{r}$.

$$\displaystyle \frac{1}{z} = \frac{r\left(\cos(-\theta) + i\sin(-\theta)\right)}{r^2} = \frac{1}{r}\left(\cos(-\theta) + i\sin(-\theta)\right)$$

Check­ing the answer#

Mul­ti­ply $z$ by the claimed $\displaystyle \tfrac{1}{z}$: $\displaystyle r \cdot \tfrac{1}{r} = 1$, and $(\cos\theta + i\sin\theta)(\cos\theta - i\sin\theta) = \cos^2\theta + \sin^2\theta = 1$. The prod­uct is $1$, as it should be.

Answer#

$\bar z = r\left(\cos(-\theta) + i\sin(-\theta)\right)$ and $\displaystyle \frac{1}{z} = \frac{1}{r}\left(\cos(-\theta) + i\sin(-\theta)\right)$.

Ques­tion 9: A locus described by a mod­u­lus#

The prob­lem#

Describe the set of points $z$ with $\lvert z - 2i \rvert = 3$.

Under­stand­ing the prob­lem#

You must find every point $z$ on the Argand plane that sat­is­fies the con­di­tion, and describe the shape they form. $2i$ is the point $(0, 2)$.

The idea#

From the les­son, $\lvert z_1 - z_2 \rvert$ is the dis­tance between two points. So $\lvert z - 2i \rvert = 3$ says "the dis­tance from $z$ to $(0, 2)$ is $3$". We con­firm this alge­braically by writ­ing $z = x + iy$.

Step-by-step solu­tion#

Step 1. Let $z = x + iy$ and group real and imag­i­nary parts.

$$z - 2i = x + i(y - 2)$$

Step 2. Write the mod­u­lus using $\lvert a + ib \rvert = \sqrt{a^2 + b^2}$.

$$\sqrt{x^2 + (y - 2)^2} = 3$$

Step 3. Square both sides (both sides are non-neg­a­tive, so noth­ing is lost).

$$x^2 + (y - 2)^2 = 9$$

Step 4. Recog­nise the shape. $(x - h)^2 + (y - k)^2 = R^2$ is a cir­cle with cen­tre $(h, k)$ and radius $R$. Here $h = 0$, $k = 2$, $R = 3$.

Check­ing the answer#

Test a point: $z = 5i$, i.e. $(0, 5)$. Then $\lvert 5i - 2i \rvert = \lvert 3i \rvert = 3$. And $z = 3 + 2i$: $\lvert 3 \rvert = 3$. Both lie on the cir­cle with cen­tre $(0, 2)$ and radius $3$.

Answer#

The set is the cir­cle with cen­tre $(0, 2)$ (the point $2i$) and radius $3$, i.e. $x^2 + (y - 2)^2 = 9$.

Ques­tion 10: Prov­ing a num­ber is purely imag­i­nary#

The prob­lem#

If $\lvert z + 1 \rvert = \lvert z - 1 \rvert$, show that $z$ is purely imag­i­nary.

Under­stand­ing the prob­lem#

You are told that $z$ sat­is­fies one equa­tion involv­ing mod­uli. You must prove that its real part is zero, so that $z = iy$ for some real $y$ (this includes $z = 0$, where $y = 0$).

The idea#

Write $z = x + iy$, turn each mod­u­lus into a square root, square both sides, and sim­plify until only a con­di­tion on $x$ is left. There is also a geo­met­ric read­ing: $\lvert z + 1 \rvert = \lvert z - (-1) \rvert$ is the dis­tance from $z$ to $-1$, and $\lvert z - 1 \rvert$ is the dis­tance from $z$ to $1$.

Step-by-step solu­tion#

Step 1. Let $z = x + iy$ with $x, y$ real. Then

$$z + 1 = (x + 1) + iy, \qquad z - 1 = (x - 1) + iy$$

Step 2. The two mod­uli are equal, so their squares are equal.

$$(x + 1)^2 + y^2 = (x - 1)^2 + y^2$$

Step 3. Can­cel $y^2$ from both sides and expand.

$$\begin{aligned} x^2 + 2x + 1 &= x^2 - 2x + 1 \\ 4x &= 0 \\ x &= 0 \end{aligned}$$

Step 4. Con­clude. The real part of $z$ is $0$, so $z = iy$: $z$ is purely imag­i­nary.

Step 5 (geo­met­ric view). $z$ is equally far from the points $-1$ and $1$. Points equidis­tant from two fixed points lie on the per­pen­dic­u­lar bisec­tor of the seg­ment join­ing them. The seg­ment from $(-1, 0)$ to $(1, 0)$ has mid­point $O$ and lies on the real axis, so its per­pen­dic­u­lar bisec­tor is the imag­i­nary axis.

Check­ing the answer#

Take $z = 2i$: $\lvert 2i + 1 \rvert = \sqrt{5}$ and $\lvert 2i - 1 \rvert = \sqrt{5}$. Equal, as expected. Take a non-imag­i­nary $z = 1 + i$: $\lvert 2 + i \rvert = \sqrt{5}$ but $\lvert i \rvert = 1$. Not equal, so it does not sat­isfy the con­di­tion.

Answer#

$\operatorname{Re} z = 0$, so $z$ lies on the imag­i­nary axis and is purely imag­i­nary.