Com­plex num­bers as points

So far com­plex num­bers have been sym­bols on paper. Here is a pic­ture that makes them far eas­ier to under­stand: think of a+iba + ib as the point (a,b)(a, b) in the plane. Seen this way, every com­plex num­ber has a dis­tance from the ori­gin (its mod­u­lus) and a direc­tion (its argu­ment), and mul­ti­ply­ing turns out to be a mix of stretch­ing and turn­ing. We will draw the Argand plane, write num­bers in polar form, and finally solve those qua­dratic equa­tions whose dis­crim­i­nant is neg­a­tive, the ones you were told had "no solu­tion" in ear­lier classes.

The Argand plane

Plot z=a+ibz = a + ib at the point (a,b)(a, b). The hor­i­zon­tal axis is now called the real axis and the ver­ti­cal one the imag­i­nary axis. Many alge­braic ideas become sim­ple geom­e­try. The con­ju­gate zˉ\bar z is the mir­ror image of zz in the real axis, and −z-z is its reflec­tion through the ori­gin. The mod­u­lus ∣z∣\lvert z \rvert is just the dis­tance from OO, and ∣z1−z2∣\lvert z_1 - z_2 \rvert is the dis­tance between two points.

Argand plane with the point z = a + ib in the first quadrant, its conjugate a - ib reflected in the real axis, and -z reflected through the origin O; the segment from O to z is marked |z|.
A com­plex num­ber, its con­ju­gate and its neg­a­tive

Polar form

Instead of say­ing how far across and how far up a point is, we can say how far away it is and in which direc­tion. If z=x+iyz = x + iy has mod­u­lus r=∣z∣r = \lvert z \rvert and makes an angle θ\theta with the pos­i­tive real axis, then x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta and

z=r(cos⁡θ+isin⁡θ).z = r(\cos\theta + i\sin\theta).

The angle is not unique, since adding a full turn lands on the same point. So we pick the prin­ci­pal argu­ment θ=arg⁡z\theta = \arg z in (−π,π](-\pi, \pi]. The safest method is in two steps. First find the ref­er­ence angle α=tan⁡−1∣yx∣\displaystyle \alpha = \tan^{-1}\left\lvert \tfrac{y}{x} \right\rvert, ignor­ing signs. Then place it accord­ing to the quad­rant of the point: in I it is α\alpha; in II, π−α\pi - \alpha; in III, −(π−α)-(\pi - \alpha); in IV, −α-\alpha. A quick sketch of the point pre­vents most mis­takes here.

Exam­ple 1. For z=−1+i3z = -1 + i\sqrt{3} we get r=2r = 2 and ref­er­ence angle π3\displaystyle \tfrac{\pi}{3}. The point is in quad­rant II, so θ=2π3\displaystyle \theta = \tfrac{2\pi}{3}, and z=2(cos⁡2π3+isin⁡2π3)\displaystyle z = 2\left(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}\right).

Argand plane showing the point -1 + i root 3 at distance r = 2 from the origin; the angle from the positive real axis is 2 pi/3 and the reference angle to the negative real axis is pi/3.
Exam­ple 1: mod­u­lus 2 and argu­ment 2 pi/3

Exam­ple 2. For z=−3−3iz = -3 - 3i, in quad­rant III, r=32r = 3\sqrt{2} and θ=−3π4\displaystyle \theta = -\tfrac{3\pi}{4}.

Exam­ple 3. Points on the axes need no for­mula. For z=−4z = -4, r=4r = 4 and θ=π\theta = \pi. For z=5iz = 5i, r=5r = 5 and θ=π2\displaystyle \theta = \tfrac{\pi}{2}.

Exam­ple 4. For 1+7i(2−i)2\displaystyle \frac{1 + 7i}{(2 - i)^2}, first sim­plify: (2−i)2=3−4i(2 - i)^2 = 3 - 4i, and (1+7i)(3+4i)25=−25+25i25=−1+i=2(cos⁡3π4+isin⁡3π4)\displaystyle \tfrac{(1 + 7i)(3 + 4i)}{25} = \tfrac{-25 + 25i}{25} = -1 + i = \sqrt{2}\left(\cos\tfrac{3\pi}{4} + i\sin\tfrac{3\pi}{4}\right).

Qua­dratic equa­tions with neg­a­tive dis­crim­i­nant

For ax2+bx+c=0ax^2 + bx + c = 0 with real coef­fi­cients and D=b2−4ac<0D = b^2 - 4ac < 0:

x=−b±i4ac−b22a.\displaystyle x = \frac{-b \pm i\sqrt{4ac - b^2}}{2a}.

Notice that the two roots always come as a pair of com­plex con­ju­gates. There is a big­ger story here too. The fun­da­men­tal the­o­rem of alge­bra says that a poly­no­mial equa­tion of degree nn has nn roots once we count in the com­plex num­bers. So with com­plex num­bers, no poly­no­mial equa­tion is ever left with­out a solu­tion.

Exam­ple 5. For x2−4x+13=0x^2 - 4x + 13 = 0, D=−36D = -36, so x=4±6i2=2±3i\displaystyle x = \tfrac{4 \pm 6i}{2} = 2 \pm 3i.

Argand plane with the two roots 2 + 3i and 2 - 3i plotted at (2, 3) and (2, -3), joined by a dashed vertical line: they are reflections of each other in the real axis.
The roots 2 + 3i and 2 - 3i are com­plex con­ju­gates

Exam­ple 6. For 3x2+2x+1=03x^2 + 2x + 1 = 0, D=−8D = -8, so x=−2±22i6=−1±2i3\displaystyle x = \tfrac{-2 \pm 2\sqrt{2} i}{6} = \tfrac{-1 \pm \sqrt{2} i}{3}.

Exam­ple 7. Surds in the coef­fi­cients change noth­ing. For 2x2+x+2=0\sqrt{2}x^2 + x + \sqrt{2} = 0, D=1−8=−7D = 1 - 8 = -7, and x=−1±7i22\displaystyle x = \tfrac{-1 \pm \sqrt{7}i}{2\sqrt{2}}.

Prac­tice ques­tions

  1. Plot 3+2i3 + 2i, its con­ju­gate and its neg­a­tive on the Argand plane.
  2. Find the dis­tance between 1+2i1 + 2i and 4−2i4 - 2i.
  3. Find the mod­u­lus and prin­ci­pal argu­ment: 1−i1 - i; −3+i-\sqrt{3} + i; −2i-2i; −3−i-\sqrt{3} - i.
  4. Write in polar form: 3+i\sqrt{3} + i; −1−i-1 - i; 66; 1+3i1−2i\displaystyle \frac{1 + 3i}{1 - 2i}.
  5. Solve: x2+9=0x^2 + 9 = 0; x2+2x+5=0x^2 + 2x + 5 = 0; 2x2−3x+4=02x^2 - 3x + 4 = 0; x2+x3+1=0\displaystyle x^2 + \tfrac{x}{\sqrt{3}} + 1 = 0.
  6. Show that the roots of x2−6x+10=0x^2 - 6x + 10 = 0 are con­ju­gates and find their prod­uct.
  7. Find a qua­dratic with real coef­fi­cients hav­ing 3−2i3 - 2i as a root.
  8. If z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta), write zˉ\bar z and 1z\displaystyle \tfrac{1}{z} in polar form.
  9. Describe the set of points zz with ∣z−2i∣=3\lvert z - 2i \rvert = 3.
  10. If ∣z+1∣=∣z−1∣\lvert z + 1 \rvert = \lvert z - 1 \rvert, show that zz is purely imag­i­nary.

Answers

Show answers
  1. (3,2)(3, 2), (3,−2)(3, -2), (−3,−2)(-3, -2).
  2. ∣−3+4i∣=5\lvert -3 + 4i \rvert = 5.
  3. 2,−π4\displaystyle \sqrt{2}, -\tfrac{\pi}{4}; 2,5π6\displaystyle 2, \tfrac{5\pi}{6}; 2,−π2\displaystyle 2, -\tfrac{\pi}{2}; 2,−5π6\displaystyle 2, -\tfrac{5\pi}{6}.
  4. 2(cos⁡π6+isin⁡π6)\displaystyle 2(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}); 2(cos⁡(−3π4)+isin⁡(−3π4))\displaystyle \sqrt{2}(\cos(-\tfrac{3\pi}{4}) + i\sin(-\tfrac{3\pi}{4})); 6(cos⁡0+isin⁡0)6(\cos 0 + i\sin 0); (1+3i)(1+2i)5=−1+i=2(cos⁡3π4+isin⁡3π4)\displaystyle \tfrac{(1 + 3i)(1 + 2i)}{5} = -1 + i = \sqrt{2}(\cos\tfrac{3\pi}{4} + i\sin\tfrac{3\pi}{4}).
  5. ±3i\pm 3i; −1±2i-1 \pm 2i; 3±23i4\displaystyle \tfrac{3 \pm \sqrt{23}i}{4}; −1±11i23\displaystyle \tfrac{-1 \pm \sqrt{11} i}{2\sqrt{3}}.
  6. 3±i3 \pm i; prod­uct 1010.
  7. x2−6x+13=0x^2 - 6x + 13 = 0.
  8. r(cos⁡(−θ)+isin⁡(−θ))r(\cos(-\theta) + i\sin(-\theta)); 1r(cos⁡(−θ)+isin⁡(−θ))\displaystyle \tfrac{1}{r}(\cos(-\theta) + i\sin(-\theta)).
  9. The cir­cle of radius 33 cen­tred at (0,2)(0, 2).
  10. zz is the same dis­tance from −1-1 and 11, so it lies on the imag­i­nary axis. Alge­braically, with z=x+iyz = x + iy, (x+1)2=(x−1)2(x + 1)^2 = (x - 1)^2 gives x=0x = 0.