3.9 - Argand diagrams and polar form

3.9 - Argand diagrams and polar form

A complex number can be located by two coordinates or by a distance and a direction. You will use an Argand diagram to connect these descriptions, interpret addition and subtraction as displacements, and convert between Cartesian, polar and exponential notation.

The Argand plane

Where should 2+i2+i be placed in a picture? Give the real part its own horizontal axis and the imaginary part its own vertical axis. The point is (2,1)(2,1): two units right and one unit up.

Argand diagram

An Argand diagram represents z=x+iyz=x+iy by the point with coordinates (x,y)(x,y). The horizontal axis is labelled Re⁡\operatorname{Re}, the vertical axis Im⁡\operatorname{Im}. Use the same scale on both axes when interpreting lengths and angles.

The vertical coordinate is a real number, even though it records an imaginary component. A point labelled 3i3i lies at height 33, on the imaginary axis; it is not a third coordinate.

An arrow from AA, representing aa, to BB, representing bb, has horizontal and vertical changes given by b−ab-a. An arrow from the origin to b−ab-a has the same length and direction. This is the geometric meaning of complex subtraction.

Worked example 1: addition as two moves

Question. Points A,BA,B represent u=2+iu=2+i and v=1+2iv=1+2i. Find u+vu+v and describe its position using the two moves from the origin.

Plan. Add the horizontal and vertical components, then interpret the result as a point.

u+v=(2+1)+i(1+2)=3+3i.u+v=(2+1)+i(1+2)=3+3i.

Move 22 right and 11 up to AA. From AA, move another 11 right and 22 up. The endpoint is C=(3,3)C=(3,3).

Answer. u+v=3+3iu+v=3+3i, represented by C(3,3)C(3,3).

Check. The arrow from AA to CC has change (1,2)(1,2), exactly the coordinates of vv. Reversing the order of the two moves reaches the same endpoint.

Keep the two numbers but replace addition by subtraction. The order of the subtraction now determines the arrow's direction.

Worked example 2: subtraction as a directed difference

Question. For the same u=2+iu=2+i and v=1+2iv=1+2i, find u−vu-v. State the relationship between the arrow from BB to AA and the arrow from the origin to the point representing u−vu-v.

Plan. Subtract coordinates in the order “endpoint minus start”.

u−v=(2−1)+i(1−2)=1−i.u-v=(2-1)+i(1-2)=1-i.

From B(1,2)B(1,2) to A(2,1)A(2,1), the change is 11 right and 11 down.

Answer. u−v=1−iu-v=1-i; both arrows have equal length and the same direction.

Check. v+(u−v)=(1+2i)+(1−i)=2+i=uv+(u-v)=(1+2i)+(1-i)=2+i=u. Using v−uv-u instead would reverse the arrow.

The copied arrows in the picture explain why addition uses a parallelogram and why subtraction uses the displacement between the original points.

[DIAGRAM: asset_name: Addition and directed subtraction; asset_slug: argand-add-subtract; caption: Addition joins two moves, while u minus v is the displacement from v to u; description: Two equal-scale Argand panels show u at 2 comma 1 and v at 1 comma 2. The left shows origin to u to u plus v at 3 comma 3, with a pale parallelogram. The right shows the arrow from v to u and an equal parallel arrow from the origin to u minus v at 1 comma minus 1.]
Addition joins two moves, while u minus v is the displacement from v to u

Use the endpoint-minus-start order with new points.

Modulus and argument

Coordinates describe a point by horizontal and vertical changes. A second description uses the length of the line from the origin and its direction from the positive real axis.

Modulus and argument

The modulus of z=x+iyz=x+iy is its distance from the origin:

∣z∣=x2+y2.|z|=\sqrt{x^2+y^2}.

For z≠0z\ne0, an argument is an angle from the positive real axis to the ray through zz, measured anticlockwise when positive. We normally choose

−π<arg⁡z≤π.-\pi<\arg z\le\pi.

These complex-number definitions are not listed in MF19.

Pythagoras gives the modulus; it is always non-negative, and only zero has modulus zero. The angle is not defined at zero because there is no ray from the origin to itself.

Use a reference triangle and then choose the quadrant. A calculator's tan⁡−1(y/x)\tan^{-1}(y/x) alone cannot distinguish opposite directions: both (1,1)(1,1) and (−1,−1)(-1,-1) give y/x=1y/x=1.

Worked example 3: above the negative real axis

Question. Find the modulus and argument of z=−3+iz=-\sqrt3+i, giving the argument in (−π,π](-\pi,\pi].

Plan. Find the distance by Pythagoras, then use the quadrant-II reference angle.

∣z∣=(−3)2+12=2.|z|=\sqrt{(-\sqrt3)^2+1^2}=2.

The reference angle satisfies tan⁡α=1/3\tan\alpha=1/\sqrt3, so α=π/6\alpha=\pi/6. The point is above the negative real axis:

arg⁡z=π−π6=5π6.\arg z=\pi-\frac\pi6=\frac{5\pi}{6}.

Answer. Modulus 22, argument 5π/65\pi/6.

Check. 2cos⁡(5π/6)=−32\cos(5\pi/6)=-\sqrt3 and 2sin⁡(5π/6)=12\sin(5\pi/6)=1, recovering both coordinates.

Change only the imaginary sign. The distance remains the same, but the direction is now below the real axis.

Worked example 4: below the negative real axis

Question. Find the modulus and argument of z=−3−iz=-\sqrt3-i, giving the argument in (−π,π](-\pi,\pi].

Plan. Keep the distance and reference angle, then choose the clockwise direction in the required range.

∣z∣=2,arg⁡z=−π+π6=−5π6.|z|=2,\qquad \arg z=-\pi+\frac\pi6=-\frac{5\pi}{6}.

Answer. Modulus 22, argument −5π/6-5\pi/6.

Check. Cosine is negative and sine is negative at this angle. Its positive coterminal angle 7π/67\pi/6 reaches the same point but lies outside the requested interval.

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