Hands first, words secondDrag w once around the origin, then pull it in and push it out. How does the green product move? How are the two triangles always related?
Loading figure…
Try this
Drag w onto the unit circle (∣w∣=1). Does multiplying by w change the length of z? What is the only thing it does?
Drag w onto the positive real axis. Does multiplying by w change the direction? What is the only thing it does?
For a general w: do the light-red triangle O-1-z and the green triangle O-w-zw have the same shape?
Look at the arcs: the red arc is argz, the blue arc that follows it is argw, and the green dashed arc is arg(zw). When does the “argument” reading show a warning?
The point Z corresponds to the complex number z=a+bi. The length r=∣z∣=a2+b2 of the vector OZ is the modulus. The angle θ whose initial side is the positive x-axis and whose terminal side is OZ is an argument, written Argz. There are infinitely many arguments, differing by integer multiples of 360∘; the one in [0,2π) is the principal argument, written argz. So
z=r(cosθ+isinθ)(r>0)
This is the trigonometric form of a complex number. Its whole point is multiplication:
z1z2=r1r2[cos(θ1+θ2)+isin(θ1+θ2)].
Predict first
What is (1+i)8?
Choose an answer first; then you can see the result.
✓You predicted it!✗Not what you expected — which is exactly why it is worth a pause.
1+i=2(cos45∘+isin45∘). Multiply it 8 times: the modulus becomes (2)8=16 and the argument becomes 8×45∘=360∘, back on the positive real axis. So (1+i)8=16.
Expanding (1+i)8 by brute force means an eighth-power binomial expansion; in trigonometric form one line is enough.
Loading figure…
z≈1+i; the figure draws 1,z,z2,…,z8. (∣z∣ is set to 1.4≈2.) Set ∣z∣ to 1 and look again.
Why
The product rule: moduli multiply, arguments add
Step 1 / 5
Loading figure…
Write both in trigonometric form
Let z1=r1(cosθ1+isinθ1) and z2=r2(cosθ2+isinθ2). In the figure r1=r2=1.5, θ1=30∘, θ2=60∘.
z1z2=r1r2(cosθ1+isinθ1)(cosθ2+isinθ2)
Expand as usual
Spot the angle sum formulas
Geometrically: similar triangles
The same number n times
Taking roots: regular polygons
Now solve wn=b the other way round. Let b=ρ(cosφ+isinφ) and w=s(cosα+isinα). De Moivre’s theorem gives sn=ρ and nα=φ+360∘k.
The sixth roots of −1
Step 1 / 4
Loading figure…
Write b in trigonometric form
−1=cos180∘+isin180∘: modulus 1, argument 180∘.
w6=−1=cos180∘+isin180∘
Solve for modulus and argument separately
Why there are only 6
They form a regular hexagon
Remove one condition
Break it
Counterexample lab: remove one condition
Conclusion: arg(zw)=argz+argw← still true?
When the two principal arguments add up to less than a full turn, the sum itself lies in [0∘,360∘), so it is the principal argument of the product.
Loading figure…
Drag z or w down into the third or fourth quadrant so that the two arguments add up to more than 360∘. The “argument” reading lights up a warning.
Loading figure…
200∘+250∘=450∘, but a principal value must be less than 360∘, so arg(zw)=90∘: a full turn less. The correct statement is
Arg(z1z2)=Argz1+Argz2,
an equality of sets (both sides mean “all the arguments”). With principal values you have to subtract an integer multiple of 360∘ to make them match.
Practice
Exercise 1Basic
Find the principal argument of z=−1+3i (in degrees; enter a number).
Hint · direction
Which quadrant is the point (−1,3) in?
Hint · method
Find the modulus first, then write z as r(cosθ+isinθ) and read off cosθ and sinθ.
Hint · first step
r=1+3=2, so cosθ=−21 and sinθ=23.
Full solution
Second quadrant: θ=120∘, that is, argz=32π.
Exercise 2Basic
Compute (cos20∘+isin20∘)9.
Hint · direction
This is already in trigonometric form. What is the modulus?
Hint · method
Use De Moivre’s theorem: raise the modulus to the 9th power and multiply the argument by 9.
Hint · first step
9×20∘=180∘.
Full solution
cos180∘+isin180∘=−1.
Exercise 3Advanced
Compute (1+i)10 and write it in the form a+bi.
Hint · direction
Both the modulus and the argument of 1+i are very tidy.
Hint · method
1+i=2(cos45∘+isin45∘); then use De Moivre’s theorem.
Hint · first step
Modulus: (2)10=32; argument: 10×45∘=450∘.
Full solution
450∘ has the same terminal side as 90∘, so (1+i)10=32(cos90∘+isin90∘)=32i.
Exercise 4Challenge
The four roots of z4=−16 form a quadrilateral in the complex plane. Find its area.
Hint · direction
What figure do the n roots of order n form?
Hint · method
The four roots lie on the circle with center the origin and radius 416, and neighboring roots are 90∘ apart.
Hint · first step
The radius is 2, and the four roots are 2(cos(45∘+90∘k)+isin(45∘+90∘k)), that is, ±2±2i.
Full solution
They form a square of side 22 (diagonal 4), with area (22)2=8.
The wrong step: −2(cos30∘+isin30∘) is in trigonometric form, with modulus −2 and argument 30∘.
What went wrong? Think first, then open
DiagnosisTrigonometric form requires r>0. The modulus is a length and can never be negative. The minus sign in front amounts to a further turn of 180∘.
✓
The right step: −2(cos30∘+isin30∘)=2(cos210∘+isin210∘): modulus 2, principal argument 210∘.
“Trigonometric form has three requirements: r positive, cos first, a plus sign in the middle”
Exhibit
Reading the argument of cos θ − i sin θ as θ
✗
The wrong step: z=cos40∘−isin40∘, so argz=40∘.
What went wrong? Think first, then open
DiagnosisThere is a minus sign in the middle, so this is not trigonometric form. The point (cos40∘,−sin40∘) is in the fourth quadrant; its argument is −40∘ plus a full turn.
✓
The right step: cos40∘−isin40∘=cos(−40∘)+isin(−40∘)=cos320∘+isin320∘, so argz=320∘.
“Rewrite in standard trigonometric form first, then read off the argument”
Modulus and Argument (I) ← → step · F full screen · Esc exit