Intuition The big picture (WHY this form exists)A complex number has two pieces of information: a size (how far from origin) and a direction (angle from positive real axis). The exponential form z = r e i θ z = re^{i\theta} z = r e i θ r r r θ \theta θ multiplication of complex numbers becomes addition of angles , because exponents add. That single fact is why engineers, physicists, and mathematicians love this form.
Definition Exponential (polar) formAny nonzero complex number can be written as
z = r e i θ z = r\,e^{i\theta} z = r e i θ
r = ∣ z ∣ = x 2 + y 2 ≥ 0 r = |z| = \sqrt{x^2+y^2} \ge 0 r = ∣ z ∣ = x 2 + y 2 ≥ 0 modulus (distance from origin),θ = arg ( z ) \theta = \arg(z) θ = arg ( z ) argument (angle, measured anticlockwise from the positive real axis).
Here x = r cos θ x = r\cos\theta x = r cos θ y = r sin θ y = r\sin\theta y = r sin θ z = x + i y z = x+iy z = x + i y
The heart of it is Euler's formula :
e i θ = cos θ + i sin θ e^{i\theta} = \cos\theta + i\sin\theta e i θ = cos θ + i sin θ
Intuition A shortcut "why" (no series needed)Let f ( θ ) = cos θ + i sin θ f(\theta)=\cos\theta+i\sin\theta f ( θ ) = cos θ + i sin θ f ′ ( θ ) = − sin θ + i cos θ = i ( cos θ + i sin θ ) = i f ( θ ) f'(\theta)=-\sin\theta+i\cos\theta = i(\cos\theta+i\sin\theta)=i\,f(\theta) f ′ ( θ ) = − sin θ + i cos θ = i ( cos θ + i sin θ ) = i f ( θ ) i i i f ( 0 ) = 1 f(0)=1 f ( 0 ) = 1 must be e i θ e^{i\theta} e i θ
Worked example Example 1 — Convertz = 1 + i z=1+i z = 1 + i Step 1: r = 1 2 + 1 2 = 2 r=\sqrt{1^2+1^2}=\sqrt2 r = 1 2 + 1 2 = 2 Why? Modulus is the distance x 2 + y 2 \sqrt{x^2+y^2} x 2 + y 2 Step 2: θ = arctan ( 1 / 1 ) = π 4 \theta=\arctan(1/1)=\dfrac{\pi}{4} θ = arctan ( 1/1 ) = 4 π Why? Argument is the angle; tan θ = y / x \tan\theta=y/x tan θ = y / x Answer: z = 2 e i π / 4 z=\sqrt2\,e^{i\pi/4} z = 2 e iπ /4
Worked example Example 2 — Convertz = − 1 + i 3 z=-1+i\sqrt3 z = − 1 + i 3 Step 1: r = ( − 1 ) 2 + ( 3 ) 2 = 4 = 2 r=\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt{4}=2 r = ( − 1 ) 2 + ( 3 ) 2 = 4 = 2 Step 2: Reference angle arctan ( 3 / 1 ) = π / 3 \arctan(\sqrt3/1)=\pi/3 arctan ( 3 /1 ) = π /3 Why check the quadrant? Here x < 0 , y > 0 x<0,\,y>0 x < 0 , y > 0 θ = π − π / 3 = 2 π 3 \theta=\pi-\pi/3=\dfrac{2\pi}{3} θ = π − π /3 = 3 2 π arctan ( y / x ) \arctan(y/x) arctan ( y / x ) Answer: z = 2 e i 2 π / 3 z=2\,e^{i2\pi/3} z = 2 e i 2 π /3
Worked example Example 3 — Use exponential form to multiplyCompute ( 1 + i ) ( − 1 + i 3 ) (1+i)(-1+i\sqrt3) ( 1 + i ) ( − 1 + i 3 ) Step 1: = 2 e i π / 4 ⋅ 2 e i 2 π / 3 = 2 2 e i ( π / 4 + 2 π / 3 ) =\sqrt2\,e^{i\pi/4}\cdot 2e^{i2\pi/3}=2\sqrt2\,e^{i(\pi/4+2\pi/3)} = 2 e iπ /4 ⋅ 2 e i 2 π /3 = 2 2 e i ( π /4 + 2 π /3 ) Why? Moduli multiply, angles add.
Step 2: π / 4 + 2 π / 3 = 3 π + 8 π 12 = 11 π 12 \pi/4+2\pi/3=\dfrac{3\pi+8\pi}{12}=\dfrac{11\pi}{12} π /4 + 2 π /3 = 12 3 π + 8 π = 12 11 π Answer: 2 2 e i 11 π / 12 2\sqrt2\,e^{i11\pi/12} 2 2 e i 11 π /12 ( 1 + i ) ( − 1 + i 3 ) = − 1 − 3 + i ( 3 − 1 ) (1+i)(-1+i\sqrt3)=-1-\sqrt3+i(\sqrt3-1) ( 1 + i ) ( − 1 + i 3 ) = − 1 − 3 + i ( 3 − 1 )
Worked example Example 4 — Powers made trivialFind ( 1 + i ) 8 (1+i)^{8} ( 1 + i ) 8 Step 1: 1 + i = 2 e i π / 4 1+i=\sqrt2\,e^{i\pi/4} 1 + i = 2 e iπ /4 Step 2: ( 1 + i ) 8 = ( 2 ) 8 e i 8 π / 4 = 16 e i 2 π = 16 (1+i)^8=(\sqrt2)^8 e^{i8\pi/4}=16\,e^{i2\pi}=16 ( 1 + i ) 8 = ( 2 ) 8 e i 8 π /4 = 16 e i 2 π = 16 Why so easy? Raising to a power multiplies the angle and powers the modulus — no messy binomial expansion.
θ = arctan ( y / x ) \theta=\arctan(y/x) θ = arctan ( y / x ) Why it feels right: tan θ = y / x \tan\theta = y/x tan θ = y / x looks like the whole answer.
The trap: arctan \arctan arctan ( − π / 2 , π / 2 ) (-\pi/2,\pi/2) ( − π /2 , π /2 ) − 1 + i 3 -1+i\sqrt3 − 1 + i 3 − π / 3 -\pi/3 − π /3 2 π / 3 2\pi/3 2 π /3 The fix: Always plot the point (or check signs of x , y x,y x , y π \pi π θ \theta θ
r r r Why it feels right: In real polar coordinates people sometimes allow negative radius.
The fix: By convention r = ∣ z ∣ ≥ 0 r=|z|\ge 0 r = ∣ z ∣ ≥ 0 θ \theta θ π \pi π − r e i θ = r e i ( θ + π ) -re^{i\theta}=re^{i(\theta+\pi)} − r e i θ = r e i ( θ + π )
e i θ e^{i\theta} e i θ e e e Why it feels right: e x e^{x} e x x x x The fix: For imaginary exponent, ∣ e i θ ∣ = cos 2 θ + sin 2 θ = 1 |e^{i\theta}|=\sqrt{\cos^2\theta+\sin^2\theta}=1 ∣ e i θ ∣ = cos 2 θ + sin 2 θ = 1 unit circle , never grows.
Recall Test yourself (hide the answers)
What does r r r ∣ z ∣ = x 2 + y 2 |z|=\sqrt{x^2+y^2} ∣ z ∣ = x 2 + y 2
What does θ \theta θ
Euler's formula? e i θ = cos θ + i sin θ e^{i\theta}=\cos\theta+i\sin\theta e i θ = cos θ + i sin θ
∣ e i θ ∣ = |e^{i\theta}|= ∣ e i θ ∣ = 1 1 1 Why do angles add on multiplication? Because e a e b = e a + b e^a e^b=e^{a+b} e a e b = e a + b
Recall Feynman: explain to a 12-year-oldImagine a clock hand of length r r r θ \theta θ multiply two such numbers, you stretch the hands (multiply their lengths) and add their spins. That's it — a complex number is just "a stretched, rotated arrow," and multiplying arrows means stretch-and-turn.
"Ready? Turn!" — R e^{iθ }: R eady = radius r r r Turn = θ \theta θ "COS is real, SIN is the swing (imaginary)" for e i θ = cos θ + i sin θ e^{i\theta}=\cos\theta+i\sin\theta e i θ = cos θ + i sin θ
What is the exponential form of a complex number? z = r e i θ z=re^{i\theta} z = r e i θ where
r = ∣ z ∣ r=|z| r = ∣ z ∣ and
θ = arg z \theta=\arg z θ = arg z .
State Euler's formula. e i θ = cos θ + i sin θ e^{i\theta}=\cos\theta+i\sin\theta e i θ = cos θ + i sin θ .
What is ∣ e i θ ∣ |e^{i\theta}| ∣ e i θ ∣ 1 1 1 , because
cos 2 θ + sin 2 θ = 1 \sqrt{\cos^2\theta+\sin^2\theta}=1 cos 2 θ + sin 2 θ = 1 ; it lies on the unit circle.
When multiplying r 1 e i θ 1 ⋅ r 2 e i θ 2 r_1e^{i\theta_1}\cdot r_2e^{i\theta_2} r 1 e i θ 1 ⋅ r 2 e i θ 2 Moduli multiply, arguments add:
r 1 r 2 e i ( θ 1 + θ 2 ) r_1r_2e^{i(\theta_1+\theta_2)} r 1 r 2 e i ( θ 1 + θ 2 ) .
Write z n z^n z n z n = r n e i n θ z^n=r^n e^{in\theta} z n = r n e in θ .
Convert 1 + i 1+i 1 + i 2 e i π / 4 \sqrt2\,e^{i\pi/4} 2 e iπ /4 .
Convert − 1 + i 3 -1+i\sqrt3 − 1 + i 3 2 e i 2 π / 3 2\,e^{i2\pi/3} 2 e i 2 π /3 (second quadrant!).
Why can't you use arctan ( y / x ) \arctan(y/x) arctan ( y / x ) It only outputs
( − π / 2 , π / 2 ) (-\pi/2,\pi/2) ( − π /2 , π /2 ) ; you must fix the quadrant using signs of
x , y x,y x , y .
How is e i θ = cos θ + i sin θ e^{i\theta}=\cos\theta+i\sin\theta e i θ = cos θ + i sin θ Put
u = i θ u=i\theta u = i θ into
e u e^u e u 's Taylor series; real terms form
cos θ \cos\theta cos θ , imaginary terms form
sin θ \sin\theta sin θ .
What is ( 1 + i ) 8 (1+i)^8 ( 1 + i ) 8 ( 2 ) 8 e i 2 π = 16 (\sqrt2)^8 e^{i2\pi}=16 ( 2 ) 8 e i 2 π = 16 .
Euler e^iθ = cosθ + i sinθ
Intuition Hinglish mein samjhoDekho, ek complex number ko samajhne ka sabse smart tareeka hai use ek arrow ki tarah sochna jo origin se nikalta hai. Us arrow ki do hi baatein important hain: uski length (jise hum r r r angle (jise θ \theta θ z = r e i θ z = re^{i\theta} z = r e i θ r r r θ \theta θ
Iska dil hai Euler's formula : e i θ = cos θ + i sin θ e^{i\theta}=\cos\theta+i\sin\theta e i θ = cos θ + i sin θ e x e^x e x i θ i\theta i θ cos θ \cos\theta cos θ sin θ \sin\theta sin θ ∣ e i θ ∣ |e^{i\theta}| ∣ e i θ ∣ 1 1 1 e e e
Sabse mazedaar baat: jab do complex numbers ko multiply karte ho, to unke r r r θ \theta θ add ho jaate hain, kyunki e a e b = e a + b e^a e^b = e^{a+b} e a e b = e a + b ( 1 + i ) 8 (1+i)^8 ( 1 + i ) 8 2 e i π / 4 \sqrt2\,e^{i\pi/4} 2 e iπ /4 16 16 16
Ek dhyan dene wali cheez: argument nikalte waqt sirf arctan ( y / x ) \arctan(y/x) arctan ( y / x ) x , y x,y x , y − 1 + i 3 -1+i\sqrt3 − 1 + i 3 2 π / 3 2\pi/3 2 π /3 − π / 3 -\pi/3 − π /3
