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3.5.5Complex Numbers

Polar form — r(cos θ + i sin θ) = r·cis θ

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🇮🇳 Yeh note Hinglish mein padho →

WHAT is polar form?

HOW do we derive it? (from scratch)

Start with a point z=a+biz=a+bi plotted on the Argand plane. Draw the arrow from origin OO to that point. Call its length rr and the angle it makes with the positive real axis θ\theta.

Step 1 — build a right triangle. Drop a perpendicular from the point to the real axis. Horizontal leg =a=a, vertical leg =b=b, hypotenuse =r=r. Why this step? Trigonometry only works on right triangles, so we manufacture one.

Step 2 — read off trig ratios. cos⁡θ=adjacenthyp=ar,sin⁡θ=oppositehyp=br\cos\theta = \frac{\text{adjacent}}{\text{hyp}} = \frac{a}{r}, \qquad \sin\theta = \frac{\text{opposite}}{\text{hyp}} = \frac{b}{r} Why this step? These are literally the definitions of cos and sin for the angle θ\theta.

Step 3 — solve for aa and bb. a=rcos⁡θ,b=rsin⁡θa = r\cos\theta, \qquad b = r\sin\theta Why this step? We want to replace a,ba,b by r,θr,\theta, so make them the subjects.

Step 4 — substitute back into z=a+biz=a+bi. z=rcos⁡θ+i (rsin⁡θ)=r(cos⁡θ+isin⁡θ)z = r\cos\theta + i\,(r\sin\theta) = r(\cos\theta + i\sin\theta) Done — that's polar form, derived purely from Pythagoras + trig. ∎

The quadrant trap (WHY tan⁡−1\tan^{-1} isn't enough)

The calculator's tan⁡−1(b/a)\tan^{-1}(b/a) only returns angles in (−90°,90°)(-90°,90°). But arrows can point anywhere in 360°360°. So you must look at which quadrant (a,b)(a,b) lives in and adjust.

| Quadrant | Signs (a,b)(a,b) | Argument from reference angle α=tan⁡−1∣ba∣\alpha=\tan^{-1}\left|\frac{b}{a}\right| | |---|---|---| | I | (+,+)(+,+) | θ=α\theta = \alpha | | II | (−,+)(-,+) | θ=π−α\theta = \pi - \alpha | | III | (−,−)(-,-) | θ=−(π−α)\theta = -(\pi-\alpha) or π+α\pi+\alpha | | IV | (+,−)(+,-) | θ=−α\theta = -\alpha |

The principal argument is chosen in (−π,π](-\pi, \pi].

Worked examples

Forecast-then-verify

Recall Predict before checking: what is

arg⁡(−1)\arg(-1) and arg⁡(−i)\arg(-i)? Forecast: −1-1 sits on the negative real axis; −i-i points straight down. Verify: arg⁡(−1)=π\arg(-1)=\pi (arrow points left). arg⁡(−i)=−π2\arg(-i)=-\frac\pi2 (principal value; pointing down). Modulus of both is 11. So −1=cis π-1=\text{cis}\,\pi and −i=cis(−π2)-i=\text{cis}\left(-\frac\pi2\right).

Common mistakes (steel-manned)

Flashcards

What does cis θ\text{cis}\,\theta stand for?
cos⁡θ+isin⁡θ\cos\theta + i\sin\theta
Formula for modulus of z=a+biz=a+bi?
r=a2+b2r=\sqrt{a^2+b^2}
Why can't you just use tan⁡−1(b/a)\tan^{-1}(b/a) for the argument?
It ignores the quadrant; tan⁡\tan has period π\pi, so points 180°180° apart give the same value. You must adjust using the quadrant of (a,b)(a,b).
Convert 1+i1+i to polar form.
2 cis π4\sqrt2\,\text{cis}\,\frac\pi4
Convert 2 cis 2π32\,\text{cis}\,\frac{2\pi}{3} to Cartesian.
−1+i3-1 + i\sqrt3
Can the modulus rr be negative?
No, r≥0r\ge0 always; direction is carried by θ\theta.
Rule for multiplying two complex numbers in polar form?
Multiply the moduli, add the arguments.
Argument of −i-i (principal value)?
−π2-\frac\pi2
In Quadrant II, how do you get θ\theta from reference angle α\alpha?
θ=π−α\theta=\pi-\alpha
What geometric object does z=a+biz=a+bi represent?
A point/arrow in the Argand (complex) plane.
Recall Feynman: explain to a 12-year-old

Imagine standing at the middle of a big field. A treasure is buried somewhere. One way to tell your friend where it is: "walk 3 steps East, then 4 steps North" — that's like 3+4i3+4i. Another way: "walk 5 steps, pointing in that direction" — you spin to the right angle and march. That "5 steps + a direction" is polar form. The number of steps is the modulus rr, and the direction you face is the angle θ\theta. Same treasure, two ways to describe it. Polar form is handy because if you want to spin and stretch arrows, it's super easy — just add angles and multiply lengths.

Connections

  • Argand Diagram — the plane where rr and θ\theta live.
  • Modulus and Argument — the two ingredients of polar form.
  • Multiplication and Division of Complex Numbers — where polar form pays off (add/subtract angles).
  • De Moivre's Theorem — (cis θ)n=cis nθ(\text{cis}\,\theta)^n = \text{cis}\,n\theta, built directly on this.
  • Euler's Formula — eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta, the deeper reason cis behaves like an exponent.
  • Roots of Complex Numbers — uses polar form to find nn-th roots.

Concept Map

described by length and angle

shorthand for

is length of arrow

is angle of arrow

gives ratios

solve for a and b

substitute into z

found via

must fix by

gives

makes easy

z = a + bi Cartesian

Polar form r cis theta

cos theta + i sin theta

Modulus r = sqrt of a2 + b2

Argument theta = arg z

Right triangle on Argand plane

cos = a/r, sin = b/r

a = r cos theta, b = r sin theta

tan inverse of b/a

Quadrant check

Principal argument in negative pi to pi

Multiply lengths, add angles

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, complex number z=a+biz=a+bi ko hum ek arrow (teer) ki tarah soch sakte hain jo origin se us point tak jaata hai Argand plane par. Cartesian form bolta hai "itna right, itna up jao" — yaani aa aur bb. Lekin polar form ek dusre andaaz mein bolta hai: "arrow kitna lamba hai" (that is modulus r=a2+b2r=\sqrt{a^2+b^2}) aur "arrow kis direction mein point kar raha hai" (that is argument θ\theta). Bas yahi do cheezein — length aur angle — se pura number describe ho jaata hai.

Derivation simple hai: point se real axis tak perpendicular giraao, ek right triangle ban jaata hai. Us triangle mein cos⁡θ=a/r\cos\theta = a/r aur sin⁡θ=b/r\sin\theta = b/r, isliye a=rcos⁡θa=r\cos\theta aur b=rsin⁡θb=r\sin\theta. Wapas z=a+biz=a+bi mein daalo to z=r(cos⁡θ+isin⁡θ)=r cis θz=r(\cos\theta+i\sin\theta)=r\,\text{cis}\,\theta. Yeh sab sirf Pythagoras aur basic trig se aaya, koi jaadu nahi.

Ek badi galti sabhi karte hain: argument ke liye seedha calculator par tan⁡−1(b/a)\tan^{-1}(b/a) maar dete hain. Yeh galat hai kyunki tan⁡\tan ka period π\pi hota hai, to Quadrant II aur Quadrant IV ke angles same aa jaate hain. Isliye pehle point ko plot karo, quadrant dekho, phir adjust karo (Q II mein π−α\pi-\alpha, Q IV mein −α-\alpha, etc.). Aur yaad rakho rr kabhi negative nahi hota — direction ki saari information angle θ\theta mein hoti hai.

Polar form itna important kyun? Kyunki multiplication ekdum aasaan ho jaata hai: moduli multiply karo, arguments add karo. Yahi cheez aage De Moivre's theorem aur roots nikaalne mein kaam aati hai. So polar form ko strong karo — poore chapter ka 80/20 ismein hai.

Go deeper — visual, from zero

D1FoundationsEvery symbol from zero · 1,744 wordsD2Visual walkthroughThe derivation in pictures · 2,568 wordsD3Worked examplesEvery scenario covered · 2,941 wordsD4ExercisesGraded, full solutions · 2,417 wordsD5Question bankConcept traps, quick reveals · 1,538 words

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Connections

Modulus - z - and argument arg(z)Maths · 3.5.4De Moivre's theorem — statement, proof, applicationsMaths · 3.5.10Euler's formula — e^(iθ) = cos θ + i sin θ (proof via Taylor series)Maths · 3.5.6nth roots of complex numbers — finding all n rootsMaths · 3.5.11Complex number a+bi — real part, imaginary partMaths · 3.5.2Argand plane — geometric representationMaths · 3.5.3Complex conjugate — properties, applications in divisionMaths · 3.5.9De Moivre's theorem — statement, proof, applicationsMaths · 3.5.10nth roots of complex numbers — finding all n rootsMaths · 3.5.11Roots of unity — cube roots, nth roots, geometric interpretationMaths · 3.5.12