A complex number z = x + i y z = x + iy z = x + i y two independent pieces of information : the real part x x x y y y point on a plane . So we treat x x x y y y Argand plane algebra of complex numbers into geometry of points and arrows . That's the whole trick: addition becomes vector addition, modulus becomes distance, argument becomes angle.
Definition Argand plane (complex plane)The plane in which the complex number z = x + i y z = x + iy z = x + i y point P ( x , y ) P(x, y) P ( x , y ) position vector O P ⃗ \vec{OP} O P
The horizontal axis is the real axis (points x + 0 i x + 0i x + 0 i
The vertical axis is the imaginary axis (points 0 + i y 0 + iy 0 + i y
Named after Jean-Robert Argand (also Wessel, Gauss).
WHY a plane and not a line? Real numbers fit on a line because one number = one coordinate. A complex number carries two real numbers, so it needs two axes → a plane.
Intuition What is modulus geometrically?The point P ( x , y ) P(x,y) P ( x , y ) O O O ∣ z ∣ |z| ∣ z ∣ z z z
The argument z z z arg z \arg z arg z θ \theta θ O P ⃗ \vec{OP} O P positive real axis , measured anticlockwise.
tan θ = y x \tan\theta = \frac{y}{x} tan θ = x y The principal argument arg z \arg z arg z ( − π , π ] (-\pi, \pi] ( − π , π ]
Common mistake Steel-man: "θ = tan − 1 ( y / x ) \theta = \tan^{-1}(y/x) θ = tan − 1 ( y / x ) Why it feels right: the triangle really does give tan θ = y / x \tan\theta = y/x tan θ = y / x tan − 1 \tan^{-1} tan − 1 Why it's wrong: tan − 1 \tan^{-1} tan − 1 ( − π / 2 , π / 2 ) (-\pi/2, \pi/2) ( − π /2 , π /2 ) z z z any quadrant, and tan θ = y / x \tan\theta = y/x tan θ = y / x The fix: find the reference angle α = tan − 1 ∣ y x ∣ \alpha = \tan^{-1}\left|\dfrac{y}{x}\right| α = tan − 1 x y
Quadrant
( x , y ) (x,y) ( x , y ) arg z \arg z arg z
I
+ , + +,+ + , + α \alpha α
II
− , + -,+ − , + π − α \pi - \alpha π − α
III
− , − -,- − , − − ( π − α ) -(\pi - \alpha) − ( π − α )
IV
+ , − +,- + , − − α -\alpha − α
Intuition Addition = tip-to-tail vectorsz 1 + z 2 z_1 + z_2 z 1 + z 2 x 1 + x 2 x_1+x_2 x 1 + x 2 y 1 + y 2 y_1+y_2 y 1 + y 2 z 1 + z 2 z_1+z_2 z 1 + z 2 diagonal of the parallelogram built on O P 1 ⃗ , O P 2 ⃗ \vec{OP_1}, \vec{OP_2} O P 1 , O P 2
Intuition Distance between two numbers∣ z 1 − z 2 ∣ |z_1 - z_2| ∣ z 1 − z 2 ∣ straight-line distance between the points P 1 P_1 P 1 P 2 P_2 P 2 ∣ z − z 0 ∣ = r |z - z_0| = r ∣ z − z 0 ∣ = r circle of radius r r r z 0 z_0 z 0
Intuition Conjugate = reflectionz ˉ = x − i y \bar z = x - iy z ˉ = x − i y z z z − z -z − z 180 ∘ 180^\circ 18 0 ∘
Worked example Example 1 — plot & describez = 3 + 4 i z = 3 + 4i z = 3 + 4 i Step 1: Point is ( 3 , 4 ) (3, 4) ( 3 , 4 ) Why this step? Both parts positive, so it's above and right of origin.
Step 2: ∣ z ∣ = 3 2 + 4 2 = 25 = 5 |z| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 ∣ z ∣ = 3 2 + 4 2 = 25 = 5 Why? Pythagoras on legs 3 , 4 3,4 3 , 4 Step 3: α = tan − 1 ( 4 / 3 ) ≈ 53.13 ∘ \alpha = \tan^{-1}(4/3) \approx 53.13^\circ α = tan − 1 ( 4/3 ) ≈ 53.1 3 ∘ arg z = 53.13 ∘ \arg z = 53.13^\circ arg z = 53.1 3 ∘ Step 4: Polar form: z = 5 ( cos 53.13 ∘ + i sin 53.13 ∘ ) z = 5(\cos 53.13^\circ + i\sin 53.13^\circ) z = 5 ( cos 53.1 3 ∘ + i sin 53.1 3 ∘ )
Worked example Example 2 — the quadrant trap:z = − 1 − i z = -1 - i z = − 1 − i Step 1: Point ( − 1 , − 1 ) (-1,-1) ( − 1 , − 1 ) Why? Both parts negative.
Step 2: ∣ z ∣ = ( − 1 ) 2 + ( − 1 ) 2 = 2 |z| = \sqrt{(-1)^2 + (-1)^2} = \sqrt 2 ∣ z ∣ = ( − 1 ) 2 + ( − 1 ) 2 = 2 Step 3: α = tan − 1 ( 1 / 1 ) = π / 4 \alpha = \tan^{-1}(1/1) = \pi/4 α = tan − 1 ( 1/1 ) = π /4 tan − 1 ( 1 ) = π / 4 \tan^{-1}(1) = \pi/4 tan − 1 ( 1 ) = π /4 wrong sign! ).
Why the correction matters: the point is in Q III, so arg z = − ( π − π / 4 ) = − 3 π 4 \arg z = -(\pi - \pi/4) = -\tfrac{3\pi}{4} arg z = − ( π − π /4 ) = − 4 3 π Step 4: z = 2 ( cos ( − 3 π 4 ) + i sin ( − 3 π 4 ) ) z = \sqrt2\left(\cos\left(-\tfrac{3\pi}{4}\right) + i\sin\left(-\tfrac{3\pi}{4}\right)\right) z = 2 ( cos ( − 4 3 π ) + i sin ( − 4 3 π ) )
Worked example Example 3 — a locus (geometry from algebra)Describe ∣ z − ( 2 + i ) ∣ = 3 |z - (2 + i)| = 3 ∣ z − ( 2 + i ) ∣ = 3 Step 1: ∣ z − z 0 ∣ = r |z - z_0| = r ∣ z − z 0 ∣ = r z 0 = 2 + i z_0 = 2+i z 0 = 2 + i r = 3 r=3 r = 3 Why: ∣ z − z 0 ∣ |z-z_0| ∣ z − z 0 ∣ z z z z 0 z_0 z 0 Step 2: "distance from ( 2 , 1 ) (2,1) ( 2 , 1 ) 3 3 3 circle , centre ( 2 , 1 ) (2,1) ( 2 , 1 ) 3 3 3
Recall Feynman: explain to a 12-year-oldImagine a treasure map. A normal number line only lets you walk left or right. But a complex number tells you TWO things: walk right/left AND walk up/down. So it points to a spot on a flat map, not just a line. The "modulus" is how far the spot is from your starting corner (measured with a ruler straight through), and the "argument" is which direction you'd point your arm to face it. Adding two complex numbers is like taking two walks one after another — you end up where the arrows stack tip-to-tail.
"Real Runs, Imaginary Ists Up" — the Real axis runs horizontal, the Imaginary axis rises up. And MAD : M odulus = A D istance (from origin).
Recall Quick self-test (hide the answers!)
What are the two axes of the Argand plane?
Derive ∣ z ∣ |z| ∣ z ∣
Why is tan − 1 ( y / x ) \tan^{-1}(y/x) tan − 1 ( y / x ) arg z \arg z arg z
What locus is ∣ z − z 0 ∣ = r |z - z_0| = r ∣ z − z 0 ∣ = r
What does the horizontal axis of the Argand plane represent? The real part
x x x of
z = x + i y z=x+iy z = x + i y (the real axis).
What does the vertical axis represent? The imaginary part
y y y (the imaginary axis).
How is z = x + i y z=x+iy z = x + i y As the point
P ( x , y ) P(x,y) P ( x , y ) / position vector
O P ⃗ \vec{OP} O P from the origin.
Derive the modulus of z z z Right triangle with legs
x , y x,y x , y ; hypotenuse
= x 2 + y 2 = ∣ z ∣ =\sqrt{x^2+y^2}=|z| = x 2 + y 2 = ∣ z ∣ by Pythagoras.
What is the argument of z z z The anticlockwise angle
θ \theta θ that
O P ⃗ \vec{OP} O P makes with the positive real axis;
tan θ = y / x \tan\theta=y/x tan θ = y / x .
Range of the principal argument? Why can't you always use θ = tan − 1 ( y / x ) \theta=\tan^{-1}(y/x) θ = tan − 1 ( y / x ) tan − 1 \tan^{-1} tan − 1 only gives
( − π / 2 , π / 2 ) (-\pi/2,\pi/2) ( − π /2 , π /2 ) (Q I & IV); you must adjust by the actual quadrant.
Polar form of a complex number? z = r ( cos θ + i sin θ ) z=r(\cos\theta+i\sin\theta) z = r ( cos θ + i sin θ ) with
r = ∣ z ∣ r=|z| r = ∣ z ∣ ,
θ = arg z \theta=\arg z θ = arg z .
Geometric meaning of ∣ z 1 − z 2 ∣ |z_1-z_2| ∣ z 1 − z 2 ∣ Straight-line distance between points
P 1 P_1 P 1 and
P 2 P_2 P 2 .
What does ∣ z − z 0 ∣ = r |z-z_0|=r ∣ z − z 0 ∣ = r A circle of radius
r r r centred at
z 0 z_0 z 0 .
Geometric effect of conjugation z ˉ \bar z z ˉ Reflection of the point across the real axis.
For z = − 1 − i z=-1-i z = − 1 − i arg z \arg z arg z − 3 π / 4 -3\pi/4 − 3 π /4 (Quadrant III).
angle from positive real axis
tan theta equals y over x
z equals r cos theta plus i sin theta
Intuition Hinglish mein samjhoDekho, ek real number ko toh hum number line pe ek point ki tarah dikha dete hain — sirf ek hi coordinate chahiye. Lekin complex number z = x + i y z = x + iy z = x + i y x x x y y y Argand plane kehte hain — horizontal axis pe real part, vertical axis pe imaginary part. Toh z = x + i y z=x+iy z = x + i y ( x , y ) (x,y) ( x , y )
Ab yeh picture kyun kaam ki hai? Kyunki algebra ab geometry ban gayi. Origin se point tak ki distance hi modulus ∣ z ∣ = x 2 + y 2 |z|=\sqrt{x^2+y^2} ∣ z ∣ = x 2 + y 2 angle banta hai, wahi argument θ \theta θ tan θ = y / x \tan\theta = y/x tan θ = y / x
Ek important trap yaad rakho: calculator ka tan − 1 ( y / x ) \tan^{-1}(y/x) tan − 1 ( y / x ) z = − 1 − i z=-1-i z = − 1 − i − 3 π / 4 -3\pi/4 − 3 π /4 π / 4 \pi/4 π /4
Iska bada faayda: addition ban jaata hai vector addition (parallelogram ka diagonal), ∣ z 1 − z 2 ∣ |z_1-z_2| ∣ z 1 − z 2 ∣ ∣ z − z 0 ∣ = r |z-z_0|=r ∣ z − z 0 ∣ = r
