Polar Coordinates and Complex Numbers

Study Sheet

Polar Coordinates and Complex Numbers

Complex plane, trig form, DeMoivre, roots, polar points, equations, and curves

Complex Numbers: Operations and the Complex Plane

Concept
The essentials

A complex number has the form z=a+biz=a+bi where aa is the real part, bb is the imaginary part, and i2=1i^2=-1.

  • [leftmargin=5mm,itemsep=1pt]
  • Add / subtract: combine real parts and imaginary parts separately.
  • Multiply: FOIL, then replace i2=1i^2=-1.
  • Conjugate: zˉ=abi\bar z=a-bi. To divide, multiply top and bottom by the conjugate of the denominator.
  • Complex plane: plot z=a+biz=a+bi as the point (a,b)(a,b); the horizontal axis is real (Re), the vertical axis is imaginary (Im).
Example
Worked example: division

Simplify 3+2i14i\dfrac{3+2i}{1-4i}. Multiply by the conjugate 1+4i1+4i:

3+2i14i1+4i1+4i=3+12i+2i+8i212+42=3+14i817=5+14i17=517+1417i.\frac{3+2i}{1-4i}\cdot\frac{1+4i}{1+4i}=\frac{3+12i+2i+8i^2}{1^2+4^2}=\frac{3+14i-8}{17}=\frac{-5+14i}{17}=-\frac{5}{17}+\frac{14}{17}i.
Tip

Powers of ii cycle every 4: i1=i, i2=1, i3=i, i4=1i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1. To simplify ini^n, divide nn by 44 and use the remainder.

Modulus, Argument, and Trigonometric (Polar) Form

Concept
Modulus and argument

For z=a+biz=a+bi:

r=z=a2+b2tanθ=ba  (choose θ in the quadrant of (a,b)).r=|z|=\sqrt{a^2+b^2}\qquad \tan\theta=\frac{b}{a}\ \ (\text{choose }\theta\text{ in the quadrant of }(a,b)).

The trigonometric (polar) form is

z=r(cosθ+isinθ)=rcisθ,z=r(\cos\theta+i\sin\theta)=r\,\text{cis}\,\theta,

where rr is the modulus and θ\theta is the argument. To go back to rectangular form, evaluate a=rcosθa=r\cos\theta and b=rsinθb=r\sin\theta.

Example
Rectangular \to trigonometric form

Write z=3+iz=-\sqrt3+i in trig form. Here a=3, b=1a=-\sqrt3,\ b=1, so

r=(3)2+12=4=2,reference angle=tan1 ⁣13=30.r=\sqrt{(-\sqrt3)^2+1^2}=\sqrt{4}=2,\qquad \text{reference angle}=\tan^{-1}\!\frac{1}{\sqrt3}=30^\circ.

The point (3,1)(-\sqrt3,1) is in Quadrant II, so θ=18030=150\theta=180^\circ-30^\circ=150^\circ. Thus

z=2(cos150+isin150)=2cis150.z=2(\cos150^\circ+i\sin150^\circ)=2\,\text{cis}\,150^\circ.
Tip

Quadrant check. A calculator's tan1\tan^{-1} only returns angles in (90,90)(-90^\circ,90^\circ). Always sketch (a,b)(a,b) and add 180180^\circ (Quadrants II, III) as needed.

Products, Quotients, and DeMoivre's Theorem

Concept
Multiply, divide, and raise to powers

Let z1=r1cisθ1z_1=r_1\,\text{cis}\,\theta_1 and z2=r2cisθ2z_2=r_2\,\text{cis}\,\theta_2.

z1z2=r1r2cis(θ1+θ2)(multiply moduli, add arguments)z1z2=r1r2cis(θ1θ2)(divide moduli, subtract arguments)\begin{aligned} z_1 z_2&=r_1 r_2\,\text{cis}(\theta_1+\theta_2) & &\text{(multiply moduli, add arguments)}\\ \frac{z_1}{z_2}&=\frac{r_1}{r_2}\,\text{cis}(\theta_1-\theta_2) & &\text{(divide moduli, subtract arguments)} \end{aligned}

DeMoivre's Theorem: for any positive integer nn,

zn=[rcisθ]n=rncis(nθ)=rn(cosnθ+isinnθ).z^n=\big[r\,\text{cis}\,\theta\big]^n=r^{\,n}\,\text{cis}(n\theta)=r^{\,n}(\cos n\theta+i\sin n\theta).
Example
DeMoivre power (full)

Compute (1+i3)4(1+i\sqrt3)^4. First convert: r=1+3=2r=\sqrt{1+3}=2 and θ=60\theta=60^\circ (Quadrant I), so 1+i3=2cis601+i\sqrt3=2\,\text{cis}\,60^\circ. By DeMoivre,

(1+i3)4=24cis(460)=16cis240.(1+i\sqrt3)^4=2^4\,\text{cis}(4\cdot 60^\circ)=16\,\text{cis}\,240^\circ.

Convert back: 16(cos240+isin240)=16 ⁣(1232i)=883i.16(\cos240^\circ+i\sin240^\circ)=16\!\left(-\tfrac12-\tfrac{\sqrt3}{2}i\right)=-8-8\sqrt3\,i.

Tip

Add or subtract 360360^\circ from the resulting angle so your final argument is a standard angle in [0,360)[0^\circ,360^\circ) before converting back.

The nnth Roots of a Complex Number

Concept
Root formula

Every nonzero complex number z=rcisθz=r\,\text{cis}\,\theta has exactly nn distinct nnth roots:

zk=rn cis ⁣(θ+360kn),k=0,1,2,,n1.z_k=\sqrt[n]{r}\ \text{cis}\!\left(\frac{\theta+360^\circ k}{n}\right),\qquad k=0,1,2,\dots,n-1.

All nn roots lie on a circle of radius rn\sqrt[n]{r}, equally spaced 360n\dfrac{360^\circ}{n} apart.

Example
All nnth roots (full)

Find the three cube roots of 8i8i. Write 8i=8cis908i=8\,\text{cis}\,90^\circ, so r3=83=2\sqrt[3]{r}=\sqrt[3]{8}=2 and the angles are 90+360k3\dfrac{90^\circ+360^\circ k}{3} for k=0,1,2k=0,1,2:

θ=30, 150, 270.\theta=30^\circ,\ 150^\circ,\ 270^\circ.
z0=2cis30=2 ⁣(32+12i)=3+i,z1=2cis150=2 ⁣(32+12i)=3+i,z2=2cis270=2(0i)=2i.\begin{aligned} z_0&=2\,\text{cis}\,30^\circ=2\!\left(\tfrac{\sqrt3}{2}+\tfrac12 i\right)=\sqrt3+i,\\ z_1&=2\,\text{cis}\,150^\circ=2\!\left(-\tfrac{\sqrt3}{2}+\tfrac12 i\right)=-\sqrt3+i,\\ z_2&=2\,\text{cis}\,270^\circ=2(0-i)=-2i. \end{aligned}
Tip

Spacing shortcut. Find the first root (k=0k=0), then add 360n\dfrac{360^\circ}{n} repeatedly to the argument to get the rest.

Polar Coordinates: Points and Conversions

Concept
Plotting and converting points
xyAB

A polar point (r,θ)(r,\theta) is located by rotating θ\theta from the positive xx-axis, then moving rr along that ray (if r<0r<0, move backward, i.e. 180180^\circ opposite).

PolarRect:x=rcosθ,y=rsinθ.\textbf{Polar}\to\textbf{Rect:}\quad x=r\cos\theta,\quad y=r\sin\theta.
RectPolar:r=x2+y2,tanθ=yx (quadrant of (x,y)).\textbf{Rect}\to\textbf{Polar:}\quad r=\sqrt{x^2+y^2},\quad \tan\theta=\frac{y}{x}\ (\text{quadrant of }(x,y)).

Multiple representations: (r,θ)=(r,θ+360)=(r,θ+180)(r,\theta)=(r,\theta+360^\circ)=(-r,\theta+180^\circ) all name the same point.

Distance and midpoint come from the coordinates.

Example
Polar \leftrightarrow rectangular (full both ways)

Polar to rectangular: convert (6,150)(6,150^\circ).

x=6cos150=6 ⁣(32)=33,y=6sin150=6 ⁣(12)=3.x=6\cos150^\circ=6\!\left(-\tfrac{\sqrt3}{2}\right)=-3\sqrt3,\qquad y=6\sin150^\circ=6\!\left(\tfrac12\right)=3.

So (6,150)(33,3)(6,150^\circ)\to(-3\sqrt3,\,3). Rectangular to polar: convert (1,3)(-1,\sqrt3).

r=(1)2+(3)2=4=2,ref angle=tan1 ⁣31=60.r=\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt4=2,\qquad \text{ref angle}=\tan^{-1}\!\frac{\sqrt3}{1}=60^\circ.

The point is in Quadrant II, so θ=18060=120\theta=180^\circ-60^\circ=120^\circ; thus (1,3)(2,120)(-1,\sqrt3)\to(2,120^\circ).

Tip

A polar point is not unique. Adding 360360^\circ or flipping the sign of rr (with a 180180^\circ shift) gives another valid name for the same location.

Converting Equations Between Polar and Rectangular Form

Concept
Substitution toolkit

Use x=rcosθx=r\cos\theta, y=rsinθy=r\sin\theta, and r2=x2+y2r^2=x^2+y^2.

  • [leftmargin=5mm,itemsep=1pt]
  • Polar \to Rect: multiply through by rr to create r2r^2, rcosθr\cos\theta, or rsinθr\sin\theta, then substitute.
  • Rect \to Polar: replace x,yx,y; solve for rr. Useful: rcosθ=xr\cos\theta=x, rsinθ=yr\sin\theta=y, so a vertical line x=cx=c becomes r=csecθr=c\sec\theta and a horizontal line y=cy=c becomes r=ccscθr=c\csc\theta.
Example
Equation conversion (both directions)

Polar to rectangular: convert r=4sinθr=4\sin\theta. Multiply by rr: r2=4rsinθr^2=4r\sin\theta, so

x2+y2=4y  x2+(y2)2=4,x^2+y^2=4y\ \Rightarrow\ x^2+(y-2)^2=4,

a circle centered at (0,2)(0,2) with radius 22. Rectangular to polar: convert y=3y=3. Substitute y=rsinθy=r\sin\theta: rsinθ=3r\sin\theta=3, so r=3sinθ=3cscθr=\dfrac{3}{\sin\theta}=3\csc\theta.

Tip

The circle r=acosθr=a\cos\theta passes through the origin with center (a2,0)\left(\tfrac{a}{2},0\right); the circle r=asinθr=a\sin\theta has center (0,a2)\left(0,\tfrac{a}{2}\right). Both have radius a2\tfrac{|a|}{2}.

Graphs of Polar Equations

Concept
Recognize the family
  • [leftmargin=5mm,itemsep=1pt]
  • Circle: r=ar=a (center origin), r=acosθr=a\cos\theta or r=asinθr=a\sin\theta (through origin).
  • Cardioid: r=a±acosθr=a\pm a\cos\theta or r=a±asinθr=a\pm a\sin\theta (heart shape; a=ba=b).
  • Lima con: r=a±bcosθr=a\pm b\cos\theta or r=a±bsinθr=a\pm b\sin\theta. Ratio ab\tfrac{a}{b}: <1<1 inner loop, =1=1 cardioid, 1<ab<21<\tfrac{a}{b}<2 dimpled, 2\ge 2 convex.
  • Rose: r=acos(nθ)r=a\cos(n\theta) or r=asin(nθ)r=a\sin(n\theta). Petal length a|a|; nn odd n\Rightarrow n petals, nn even 2n\Rightarrow 2n petals.
  • Lemniscate: r2=a2cos2θr^2=a^2\cos2\theta or r2=a2sin2θr^2=a^2\sin2\theta (figure-eight).

Tip

Rose petal count trap: the number of petals depends on the parity of nn, not on nn itself. r=cos3θr=\cos3\theta has 33 petals; r=cos2θr=\cos2\theta has 44 petals; r=cos4θr=\cos4\theta has 88 petals.

Going Deeper: Advanced Polar & Complex Ideas

Concept
Roots of unity: sums and products

The nnth roots of unity are the solutions of zn=1z^n=1. Writing ω=cis360n\omega=\text{cis}\dfrac{360^\circ}{n}, they are the nn equally spaced points

1, ω, ω2, , ωn1,ωk=cis ⁣(360kn).1,\ \omega,\ \omega^2,\ \dots,\ \omega^{\,n-1},\qquad \omega^k=\text{cis}\!\left(\frac{360^\circ k}{n}\right).

Two identities fall out of the geometry:

  • [leftmargin=5mm,itemsep=1pt]
  • Sum is zero: k=0n1ωk=1+ω++ωn1=0\displaystyle\sum_{k=0}^{n-1}\omega^k=1+\omega+\dots+\omega^{\,n-1}=0 for n2n\ge 2 (finite geometric series with ratio ω1\omega\ne1 and ωn=1\omega^n=1). The roots balance about the origin.
  • Product identity: k=1n1(1ωk)=n\displaystyle\prod_{k=1}^{n-1}\bigl(1-\omega^k\bigr)=n. (Factor zn1=k=0n1(zωk)z^n-1=\prod_{k=0}^{n-1}(z-\omega^k), divide by z1z-1, then let z1z\to1.)
Example
Worked example: all fifth roots of unity, and their sum

Solve z5=1z^5=1. Here r=1r=1 and θ=0\theta=0^\circ, so 15=1\sqrt[5]{1}=1 and the arguments are 0+360k5=72k\dfrac{0^\circ+360^\circ k}{5}=72^\circ k for k=0,1,2,3,4k=0,1,2,3,4. All five roots:

z0=cis0=1,z1=cis72,z2=cis144,z3=cis216,z4=cis288.\begin{aligned} z_0&=\text{cis}\,0^\circ=1, & z_1&=\text{cis}\,72^\circ, & z_2&=\text{cis}\,144^\circ,\\ z_3&=\text{cis}\,216^\circ, & z_4&=\text{cis}\,288^\circ. && \end{aligned}

They sit at the vertices of a regular pentagon on the unit circle. Their sum pairs conjugates:

1+2cos72+2cos144=1+2(0.309)+2(0.809)=0,1+2\cos72^\circ+2\cos144^\circ=1+2(0.309)+2(-0.809)=0,

confirming k=04zk=0\sum_{k=0}^{4}z_k=0.

Concept
The identity zn+zn=2cosnθz^n+z^{-n}=2\cos n\theta

If z=cisθz=\text{cis}\,\theta lies on the unit circle then 1z=z1=cis(θ)=cosθisinθ\dfrac1z=z^{-1}=\text{cis}(-\theta)=\cos\theta-i\sin\theta. By DeMoivre, zn=cisnθz^n=\text{cis}\,n\theta and zn=cis(nθ)z^{-n}=\text{cis}(-n\theta), so adding and subtracting gives

zn+zn=2cosnθ,znzn=2isinnθ.z^n+z^{-n}=2\cos n\theta,\qquad z^n-z^{-n}=2i\sin n\theta.

The special case z+1z=2cosθz+\dfrac1z=2\cos\theta is the workhorse: raising it to a power and expanding lets you rewrite cosmθ\cos^m\theta as a sum of terms coskθ\cos k\theta (Fourier/power-reduction identities).

Example
Worked example: DeMoivre proves the triple-angle formulas

Expand (cosθ+isinθ)3(\cos\theta+i\sin\theta)^3 two ways. By DeMoivre it equals cos3θ+isin3θ\cos3\theta+i\sin3\theta. By the binomial theorem (writing c=cosθ, s=sinθc=\cos\theta,\ s=\sin\theta),

(c+is)3=c3+3c2(is)+3c(is)2+(is)3=c33cs2+i(3c2ss3).(c+is)^3=c^3+3c^2(is)+3c(is)^2+(is)^3=c^3-3cs^2+i\,(3c^2s-s^3).

Matching real and imaginary parts:

cos3θ=cos3θ3cosθsin2θ=4cos3θ3cosθ,sin3θ=3cos2θsinθsin3θ=3sinθ4sin3θ,\begin{aligned} \cos3\theta&=\cos^3\theta-3\cos\theta\sin^2\theta=4\cos^3\theta-3\cos\theta,\\ \sin3\theta&=3\cos^2\theta\sin\theta-\sin^3\theta=3\sin\theta-4\sin^3\theta, \end{aligned}

using sin2θ=1cos2θ\sin^2\theta=1-\cos^2\theta (and cos2θ=1sin2θ\cos^2\theta=1-\sin^2\theta) to simplify.

Reminder — The binomial theorem:(x+y)n=k=0n(nk)xnkyk(x+y)^{n}=\sum_{k=0}^{n}\binom{n}{k}x^{\,n-k}y^{k}
Concept
Loci in the complex plane: the Apollonius circle

Treat z=x+iyz=x+iy as the point (x,y)(x,y) and read za|z-a| as the distance from zz to the point aa.

  • [leftmargin=5mm,itemsep=1pt]
  • za=k|z-a|=k is a circle of radius kk centered at aa.
  • za=zb|z-a|=|z-b| is the perpendicular bisector of the segment abab.
  • za=kzb|z-a|=k|z-b| with k1k\ne1 is the Apollonius circle: the set of points whose distances to aa and bb are in a fixed ratio kk.

To find it, square both sides and expand with za2=(xa1)2+(ya2)2|z-a|^2=(x-a_1)^2+(y-a_2)^2; the x2x^2 and y2y^2 terms survive with equal coefficients, so the result is always a circle.

Example
Worked example: the locus z1=12z4|z-1|=\tfrac12|z-4|

Square: z12=14z42|z-1|^2=\tfrac14|z-4|^2, i.e. (x1)2+y2=14[(x4)2+y2](x-1)^2+y^2=\tfrac14\bigl[(x-4)^2+y^2\bigr]. Multiply by 44:

4(x1)2+4y2=(x4)2+y2.4(x-1)^2+4y^2=(x-4)^2+y^2.

Expand: 4x28x+4+4y2=x28x+16+y24x^2-8x+4+4y^2=x^2-8x+16+y^2. The 8x-8x terms cancel, leaving

3x2+3y2=12  x2+y2=4,3x^2+3y^2=12\ \Longrightarrow\ x^2+y^2=4,

a circle of radius 22 centered at the origin (drawn above).

Tip

Sum of the roots of zn=wz^n=w. The equation znw=0z^n-w=0 has no zn1z^{\,n-1} term, so by Vieta's formulas the nn roots always sum to 00 (for n2n\ge2), and their product is (1)n+1w(-1)^{n+1}w. Geometrically the roots are nn points equally spaced on a circle of radius wn\sqrt[n]{|w|}, so their centroid is the origin.

Reminder — Vieta's formulas:r1+r2=ba,r1r2=car_1+r_2=-\frac{b}{a},\qquad r_1 r_2=\frac{c}{a}
Concept
Area enclosed by a polar curve

A thin wedge of angle dθd\theta at radius rr is a circular sector of area 12r2dθ\tfrac12 r^2\,d\theta. Summing wedges, the area swept by r=f(θ)r=f(\theta) from θ=α\theta=\alpha to θ=β\theta=\beta is

A=12αβr2dθ=12αβ[f(θ)]2dθ.A=\frac12\int_{\alpha}^{\beta} r^2\,d\theta=\frac12\int_{\alpha}^{\beta}\bigl[f(\theta)\bigr]^2\,d\theta.

Choosing limits: for one rose petal, integrate between the two consecutive θ\theta where r=0r=0; for a full cardioid, use 00 to 360360^\circ once around. For the area between two curves, subtract: A=12(rout2rin2)dθA=\tfrac12\int(r_{\text{out}}^2-r_{\text{in}}^2)\,d\theta.

Example
Worked example: area of one petal and a curve intersection

Area of one petal of r=2cos2θr=2\cos2\theta. A petal runs from θ=45\theta=-45^\circ to θ=45\theta=45^\circ (where r=0r=0):

A=12π/4π/4(2cos2θ)2dθ=π/4π/4(1+cos4θ)dθ=[θ+sin4θ4]π/4π/4=π2,A=\frac12\int_{-\pi/4}^{\pi/4}(2\cos2\theta)^2\,d\theta =\int_{-\pi/4}^{\pi/4}(1+\cos4\theta)\,d\theta =\Bigl[\theta+\tfrac{\sin4\theta}{4}\Bigr]_{-\pi/4}^{\pi/4}=\frac{\pi}{2},

using 4cos22θ=2(1+cos4θ)4\cos^2 2\theta=2(1+\cos4\theta). Intersection of r=1+cosθr=1+\cos\theta and r=3cosθr=3\cos\theta. Set the radii equal:

1+cosθ=3cosθ  2cosθ=1  cosθ=12  θ=60,300,1+\cos\theta=3\cos\theta\ \Rightarrow\ 2\cos\theta=1\ \Rightarrow\ \cos\theta=\tfrac12\ \Rightarrow\ \theta=60^\circ,\,300^\circ,

giving r=32r=\tfrac32: the points (32,60)\bigl(\tfrac32,60^\circ\bigr) and (32,300)\bigl(\tfrac32,300^\circ\bigr). Also check the pole separately: the cardioid reaches r=0r=0 at θ=180\theta=180^\circ and the circle reaches r=0r=0 at θ=90\theta=90^\circ, so both pass through the origin even though no single θ\theta solves the equation there.

Tip

Polar intersection trap. Setting r1(θ)=r2(θ)r_1(\theta)=r_2(\theta) can miss intersections, because the same point has many names (r,θ)(r,\theta). Always test the pole and try replacing (r,θ)(r,\theta) with (r,θ+180)(-r,\theta+180^\circ) or (r,θ+360)(r,\theta+360^\circ) in one equation before solving.

Formulas, Proofs & Tips

Tip
Polar form and De Moivre’s theorem
z=r(cosθ+isinθ),zn=rn(cosnθ+isinnθ)z=r(\cos\theta+i\sin\theta),\qquad z^{n}=r^{n}\big(\cos n\theta+i\sin n\theta\big)

What it means. A complex number is a length and a direction; powering it powers the length and multiplies the angle.

Example. (1+i)4(1+i)^4: here r=2r=\sqrt2, θ=45\theta=45^\circ, so r4=4r^4=4 and 4θ=1804\theta=180^\circ, giving 4-4.

Why it works. Multiplying two complex numbers in polar form and applying the sine and cosine sum formulas produces r1r2r_1r_2 with angle θ1+θ2\theta_1+\theta_2 — multiplication adds angles. Repeating nn times gives De Moivre.

Tip. r=x2+y2r=\sqrt{x^2+y^2} and θ=arctanyx\theta=\arctan\tfrac{y}{x} — but check the quadrant, since arctan\arctan only returns two of them.