Intuition The big pictureA complex number is just a 2D arrow on a plane. Rectangular form z = a + b i z=a+bi z = a + bi shadow on two axes (horizontal a a a b b b z = r ( cos θ + i sin θ ) z=r(\cos\theta+i\sin\theta) z = r ( cos θ + i sin θ ) same arrow by its length r r r θ \theta θ .
Add/subtract = line up arrows tip-to-tail → easiest in rectangular .Multiply/divide = scale lengths and rotate angles → easiest in polar .
The whole subtopic is: pick the form that matches the operation.
Definition Rectangular and polar form
Rectangular: z = a + b i z = a + bi z = a + bi a = Re ( z ) a=\operatorname{Re}(z) a = Re ( z ) b = Im ( z ) b=\operatorname{Im}(z) b = Im ( z ) i 2 = − 1 i^2=-1 i 2 = − 1 Modulus: r = ∣ z ∣ = a 2 + b 2 r=|z|=\sqrt{a^2+b^2} r = ∣ z ∣ = a 2 + b 2 Argument: θ = arg ( z ) \theta=\arg(z) θ = arg ( z ) cos θ = a r , sin θ = b r \cos\theta=\dfrac{a}{r},\ \sin\theta=\dfrac{b}{r} cos θ = r a , sin θ = r b Polar: z = r ( cos θ + i sin θ ) = r cis θ = r e i θ z=r(\cos\theta+i\sin\theta)=r\,\text{cis}\,\theta = re^{i\theta} z = r ( cos θ + i sin θ ) = r cis θ = r e i θ
HOW to convert both ways:
a = r cos θ , b = r sin θ , r = a 2 + b 2 , θ = atan2 ( b , a ) . a=r\cos\theta,\qquad b=r\sin\theta,\qquad r=\sqrt{a^2+b^2},\qquad \theta=\operatorname{atan2}(b,a). a = r cos θ , b = r sin θ , r = a 2 + b 2 , θ = atan2 ( b , a ) .
θ = arctan ( b / a ) \theta = \arctan(b/a) θ = arctan ( b / a ) Why it feels right: tan θ = b / a \tan\theta = b/a tan θ = b / a The trap: plain arctan \arctan arctan ( − 90 ∘ , 90 ∘ ) (-90^\circ,90^\circ) ( − 9 0 ∘ , 9 0 ∘ ) z = − 1 − i z=-1-i z = − 1 − i a , b < 0 a,b<0 a , b < 0 arctan ( 1 ) = 45 ∘ \arctan(1)=45^\circ arctan ( 1 ) = 4 5 ∘ 225 ∘ 225^\circ 22 5 ∘ − 135 ∘ -135^\circ − 13 5 ∘ Fix: check the quadrant from the signs of a a a b b b ; add ± 180 ∘ \pm180^\circ ± 18 0 ∘ a < 0 a<0 a < 0 atan2 does for you.
Intuition Adding arrows = adding componentsTwo arrows added tip-to-tail: the total horizontal reach is the sum of horizontals, total vertical reach is the sum of verticals. So you add the parts , exactly like vectors.
( 3 + 2 i ) + ( 1 − 5 i ) (3+2i)+(1-5i) ( 3 + 2 i ) + ( 1 − 5 i ) = ( 3 + 1 ) + ( 2 − 5 ) i = 4 − 3 i = (3+1)+(2-5)i = 4-3i = ( 3 + 1 ) + ( 2 − 5 ) i = 4 − 3 i Why this step? Real parts 3 , 1 3,1 3 , 1 2 , − 5 2,-5 2 , − 5
Common mistake Adding in polar formWhy it feels right: polar is "the good form" for everything else. Trap: you cannot add moduli or angles: r 1 + r 2 r_1+r_2 r 1 + r 2 θ 1 + θ 2 \theta_1+\theta_2 θ 1 + θ 2 Fix: convert to rectangular, add, convert back if needed.
( 1 + i ) ( 1 + i ) (1+i)(1+i) ( 1 + i ) ( 1 + i ) Rectangular: 1 + i + i + i 2 = 1 + 2 i − 1 = 2 i 1+i+i+i^2 = 1+2i-1 = 2i 1 + i + i + i 2 = 1 + 2 i − 1 = 2 i Polar: 1 + i 1+i 1 + i r = 2 r=\sqrt2 r = 2 θ = 45 ∘ \theta=45^\circ θ = 4 5 ∘ r = ( 2 ) 2 = 2 r=(\sqrt2)^2=2 r = ( 2 ) 2 = 2 θ = 90 ∘ \theta=90^\circ θ = 9 0 ∘ 2 ( cos 90 ∘ + i sin 90 ∘ ) = 2 i 2(\cos90^\circ+i\sin90^\circ)=2i 2 ( cos 9 0 ∘ + i sin 9 0 ∘ ) = 2 i Why this step? Both routes must agree — polar shows the geometry (double the angle, square the length).
Intuition Rationalising the denominatorWe can't "divide by an arrow," but multiplying top and bottom by the conjugate z ˉ = c − d i \bar z=c-di z ˉ = c − d i real number , because ( c + d i ) ( c − d i ) = c 2 + d 2 = ∣ z 2 ∣ 2 (c+di)(c-di)=c^2+d^2=|z_2|^2 ( c + d i ) ( c − d i ) = c 2 + d 2 = ∣ z 2 ∣ 2
3 + 2 i 1 − i \dfrac{3+2i}{1-i} 1 − i 3 + 2 i 1 + i 1+i 1 + i ( 3 + 2 i ) ( 1 + i ) ( 1 − i ) ( 1 + i ) = 3 + 3 i + 2 i + 2 i 2 1 − i 2 = ( 3 − 2 ) + 5 i 2 = 1 + 5 i 2 = 0.5 + 2.5 i . \frac{(3+2i)(1+i)}{(1-i)(1+i)}=\frac{3+3i+2i+2i^2}{1-i^2}=\frac{(3-2)+5i}{2}=\frac{1+5i}{2}=0.5+2.5i. ( 1 − i ) ( 1 + i ) ( 3 + 2 i ) ( 1 + i ) = 1 − i 2 3 + 3 i + 2 i + 2 i 2 = 2 ( 3 − 2 ) + 5 i = 2 1 + 5 i = 0.5 + 2.5 i . Why this step? Denominator 1 − i 2 = 2 1-i^2=2 1 − i 2 = 2
Common mistake Forgetting to conjugate thewhole denominator
Why it feels right: "i i i Trap: students write a + b i c + d i ⋅ 1 c − d i \frac{a+bi}{c+di}\cdot\frac{1}{c-di} c + d i a + bi ⋅ c − d i 1 numerator too — that changes the value. Fix: multiply top and bottom by c − d i c-di c − d i 1 1 1
Recall Feynman: explain to a 12-year-oldThink of each complex number as an arrow on graph paper . To add two arrows you just stack them tip-to-tail — how far right and how far up simply pile up, so add the sideways numbers and add the up numbers. To multiply , it's a different game: you stretch the arrow and spin it. The length of the answer is the two lengths multiplied, and the direction is the two angles added together. Dividing undoes that: divide the lengths and subtract the angles. That's the whole story — adding is "line them up," multiplying is "stretch and spin."
Mnemonic Which form for which job?"Add in the box, multiply on the clock."
Box = rectangular (add/subtract components). Clock = polar (multiply/divide → turn the hands , angles add/subtract).
Rectangular form of a complex number z = a + b i z=a+bi z = a + bi with
a = Re a=\operatorname{Re} a = Re ,
b = Im b=\operatorname{Im} b = Im ,
i 2 = − 1 i^2=-1 i 2 = − 1 .
Polar form of a complex number z = r ( cos θ + i sin θ ) = r e i θ z=r(\cos\theta+i\sin\theta)=re^{i\theta} z = r ( cos θ + i sin θ ) = r e i θ ,
r = ∣ z ∣ r=|z| r = ∣ z ∣ ,
θ = arg z \theta=\arg z θ = arg z .
Formula for modulus r r r a , b a,b a , b r = a 2 + b 2 r=\sqrt{a^2+b^2} r = a 2 + b 2 .
Why can't you use plain arctan ( b / a ) \arctan(b/a) arctan ( b / a ) It ignores the quadrant; use signs of
a , b a,b a , b (or atan2) and add
± 180 ∘ \pm180^\circ ± 18 0 ∘ when
a < 0 a<0 a < 0 .
Which form is best for addition and why? Rectangular — you just add real and imaginary parts independently (like vectors).
Rectangular multiplication result of ( a + b i ) ( c + d i ) (a+bi)(c+di) ( a + bi ) ( c + d i ) ( a c − b d ) + ( a d + b c ) i (ac-bd)+(ad+bc)i ( a c − b d ) + ( a d + b c ) i .
Why does a c − b d ac-bd a c − b d Because
b d i 2 = − b d bdi^2=-bd b d i 2 = − b d moves into the real part.
Polar rule for multiplication Multiply moduli, add arguments:
r 1 r 2 cis ( α + β ) r_1r_2\,\text{cis}(\alpha+\beta) r 1 r 2 cis ( α + β ) .
Which trig identities make polar multiplication work? Angle-sum identities for
cos ( α + β ) \cos(\alpha+\beta) cos ( α + β ) and
sin ( α + β ) \sin(\alpha+\beta) sin ( α + β ) .
How do you divide two complex numbers in rectangular form? Multiply numerator and denominator by the conjugate of the denominator.
Result of ( c + d i ) ( c − d i ) (c+di)(c-di) ( c + d i ) ( c − d i ) c 2 + d 2 = ∣ z ∣ 2 c^2+d^2=|z|^2 c 2 + d 2 = ∣ z ∣ 2 , a real number.
Polar rule for division Divide moduli, subtract arguments:
r 1 r 2 cis ( α − β ) \frac{r_1}{r_2}\,\text{cis}(\alpha-\beta) r 2 r 1 cis ( α − β ) .
Geometric meaning of multiplying two complex numbers Scale (stretch) length and rotate (spin) angle.
named by length and angle
multiply moduli add angles
Complex number as 2D arrow
Intuition Hinglish mein samjhoDekho, ek complex number ko ek arrow (teer) samjho jo graph paper pe bana hai. Rectangular form a + b i a+bi a + bi r cis θ r\,\text{cis}\,\theta r cis θ
Ab funda simple hai: jodna-ghatana (add/subtract) rectangular mein karo — real part real ke saath, imaginary part imaginary ke saath, bilkul vectors ki tarah tip-to-tail. Ismein polar use mat karo, kyun ki na moduli add hote hain na angles — woh galat answer dega.
Guna-bhaag (multiply/divide) polar mein maza deta hai. Multiply karte waqt lengths (moduli) multiply ho jaati hain aur angles add ho jaate hain — matlab arrow ko stretch karo aur ghumao. Divide mein ulta: lengths divide, angles subtract. Rectangular mein multiply karna ho to FOIL karke i 2 = − 1 i^2=-1 i 2 = − 1 conjugate se multiply karo taaki neeche real number bane.
Yaad rakhne ka mantra: "Add in the box, multiply on the clock." Box matlab rectangular, clock matlab polar (angle ki suiyaan ghumti hain). Bas quadrant ka dhyaan rakhna jab argument nikaalo — sirf arctan \arctan arctan 180 ∘ 180^\circ 18 0 ∘