Rectangular form z=x+iy is natural for addition/subtraction (just combine components), but multiplication, powers and roots are far easier in an alternate representation: polar form.
Polar coordinates. Superimposing polar coordinates (r,θ) — r the distance from the pole O, θ the angle from the initial line, measured counter-clockwise — onto the rectangular Argand plane gives
x=rcosθ,y=rsinθ,
so any nonzero z=x+iy can be written
z=rcosθ+irsinθ=r(cosθ+isinθ)=rcisθ.
Here r=∣z∣=x2+y2 is the modulus, and θ (found from tanθ=y/x, with the quadrant of z fixing which angle) is an argument of z, written argz. Since adding any multiple of 2π to θ gives the same point, argz has infinitely many values, all differing by 2kπ. The unique value with −π<θ≤π is the principal argument, Argz; every general argument is argz=Argz+2kπ,k∈Z. (For z=0, θ is undefined, so polar form always assumes z=0.) Conjugation flips the sign of the argument: if z has polar coordinates (r,θ), z has (r,−θ).
Argument properties (mirroring the modulus properties):
Euler's form. Euler's formula identifies the trigonometric bracket with a complex exponential,
eiθ=cosθ+isinθ,
giving the compact exponential (Euler) formz=reiθ. This form is especially convenient for multiplication (exponents add: r1eiθ1⋅r2eiθ2=r1r2ei(θ1+θ2)), and — as in de Moivre's Theorem — for computing powers and roots, since (reiθ)n=rneinθ falls straight out of the ordinary exponent law.
For each number find r=∣z∣=x2+y2, the reference angle α=tan−1∣y/x∣, then fix θ by the quadrant of (x,y), and write z=rcisθ.
Q1 ⇒θ=α.
Q4 ⇒θ=−α.
Q3 ⇒θ=α−π.
simplify first, then Q2 ⇒θ=π−α.
✓Final answer
(i) 4cis3π (ii) 23cis(−6π) (iii) 22cis(−43π) (iv) 2cis125π.
In each part we compute r=x2+y2, the reference angle α=tan−1xy, then read off the principal argument θ from the quadrant rule, mirroring Examples 2.22–2.23.
Step 1. Part (i): 2+i23. Here x=2,y=23.
r=22+(23)2=4+12=16=4.
α=tan−1223=tan−13=3π.
Since x>0,y>0, the point lies in Quadrant I, so θ=α=3π.
2+i23=4(cos3π+isin3π)=4cis3π.
Step 2. Part (ii): 3−i3. Here x=3,y=−3.
r=9+3=12=23.
α=tan−13−3=tan−131=6π.
Since x>0,y<0 (Quadrant IV), θ=−α=−6π.
3−i3=23cis(−6π).
Step 3. Part (iii): −2−i2. Here x=−2,y=−2.
r=4+4=22.
α=tan−1−2−2=tan−11=4π.
Since x<0,y<0 (Quadrant III), θ=α−π=4π−π=−43π.
−2−i2=22cis(−43π).
Step 4. Part (iv): simplify cos3π+isin3πi−1 first. The numerator is −1+i; the denominator is already in cis form, cis3π.
For −1+i: x=−1,y=1, so r=1+1=2, α=tan−1−11=4π; Quadrant II gives θ=π−α=43π. So −1+i=2cis43π.
Step 5. Part (iv): divide using the quotient rule arg(z1/z2)=argz1−argz2.