The Cartesian form z=a+bi is convenient for addition. The polar formz=r(cosθ+isinθ) is convenient for multiplication. Switching between the two is the single most important skill in this chapter.
Definitions
For any z=0 in the Argand plane, let
r=∣z∣,θ=arg(z).
Then
z=r(cosθ+isinθ).
This is the polar (or trigonometric) form of z. The pair (r,θ) are polar coordinates of the point. We sometimes write the right side as rcisθ , a shorthand for "r times cosθ+isinθ".
The argument θ is determined up to adding 2π. The principal argumentArg(z)∈(−π,π] is the canonical choice.
Conversion
Cartesian to polar. Given z=a+bi:
r=a2+b2,tanθ=ab (with quadrant adjustment).
Polar to Cartesian. Given r,θ:
a=rcosθ,b=rsinθ.
Multiplication and division in polar form
Let z1=r1(cosθ1+isinθ1) and z2=r2(cosθ2+isinθ2).
Then
z1z2=r1r2(cos(θ1+θ2)+isin(θ1+θ2)),z2z1=r2r1(cos(θ1−θ2)+isin(θ1−θ2)).
Multiply moduli, add arguments.Divide moduli, subtract arguments. This is the geometric content of complex multiplication, and the deepest reason why complex numbers are useful.
Proof of the multiplication formula
z1z2=r1r2(cosθ1+isinθ1)(cosθ2+isinθ2).
Expanding:
=r1r2[(cosθ1cosθ2−sinθ1sinθ2)+i(sinθ1cosθ2+cosθ1sinθ2)].
By the addition formulas (Chapter 3):
=r1r2[cos(θ1+θ2)+isin(θ1+θ2)].\qed
De Moivre's theorem (preview)
For any positive integer n and any complex number z=r(cosθ+isinθ):
zn=rn(cosnθ+isinnθ).
This is De Moivre's theorem. Repeated application of the multiplication formula gives the case n∈N; the formula extends to negative integers (and ultimately to all real numbers via zn=enlnz in Class XII).
Roots of unity
The n-th roots of unity are the complex solutions of zn=1. They are
zk=cosn2πk+isinn2πk,k=0,1,…,n−1.
They form a regular n-gon inscribed in the unit circle, with one vertex at 1.
Worked examples
Example 1. Express z=1+i in polar form.
r=2, argz=π/4. So z=2(cos4π+isin4π).
Example 2. Express z=−3+i in polar form.
r=3+1=2. Point (−3,1) is in quadrant II, so θ=π−π/6=5π/6. So z=2(cos65π+isin65π).
Example 3. Compute (1+i)10 using polar form.
1+i=2cis4π. So (1+i)10=(2)10cis410π=32cis25π=32cis2π=32⋅i=32i.
Example 4. Find the three cube roots of unity.
z3=1, so zk=cis32πk for k=0,1,2:
z0=1.
z1=cos32π+isin32π=−21+23i (often called ω).
z2=cos34π+isin34π=−21−23i=ω2.
Note: 1+ω+ω2=0 and ω3=1.
Example 5 (harder). Find all complex solutions of z4=16.
16=16cis0. Fourth roots: zk=161/4cis42πk=2cis2πk for k=0,1,2,3:
z0=2.
z1=2i.
z2=−2.
z3=−2i.
Try it yourself
Write z=−1−i in polar form (principal argument).
Write z=4cis3π in Cartesian form.
Compute (1−i)6 using polar form.
Find the modulus and argument of 1−i1+i.
Solve z3=−8.
Find all sixth roots of unity.
Compute (3+i)12.
Prove (cosθ+isinθ)n+(cosθ−isinθ)n=2cosnθ.
Find z with ∣z∣=2 and argz=−4π.
Show that ω=e2πi/3 satisfies ω2+ω+1=0.
Find all solutions of z2=i.
Express cos5θ as a polynomial in cosθ using De Moivre.
Pitfalls / Tricks
The argument is not unique. Always specify whether you want the principal value or "any" argument.
For z in quadrant II or III, arctan(b/a) gives the wrong sign , add or subtract π.
Polar form makes powers and roots easy; Cartesian form makes addition easy. Switch as needed.
Insight. Multiplication by cisθ is rotation by angle θ. Multiplication by r is scaling by r. Every complex multiplication is a rotation-and-scaling.