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Complex numbers and the Argand plane · Measure modulus and argument

Choose a principal argument

Plot z = x + iy. Check zero, then choose its axis or quadrant. Measure from the positive real axis: anticlockwise is positive; clockwise is negative.

  1. Check that the point has a direction

    Write z = x + iy with real x and y. The principal interval here includes π and excludes −π.

    −π<θ≤π-\pi<\theta\le\pi

    x = 0 and y = 0Stop: zero has no argument

    The zero vector has no direction. Its modulus is ∣0∣=0|0|=0, but an argument is undefined; 0/00/0 cannot supply an angle.

    At least one coordinate is nonzeroChoose the matching position case

    If either coordinate is zero, use the axis cases. If both are nonzero, use the quadrant cases. A zero real part does not make the argument undefined when the imaginary part is nonzero.

    Keep this interval throughout the calculation. The two directions π and −π reach the same negative real ray, but only π belongs to the chosen principal interval.

  2. Read axis directions directly

    Use this stage only for a nonzero point on an axis. One coordinate is zero, so an acute reference triangle is unnecessary.

    x=0x=0 or y=0y=0

    Positive real · x > 0, y = 0No turn

    θ=0\theta=0

    Negative real · x < 0, y = 0Half a turn anticlockwise

    θ=π\theta=\pi

    Use π\pi, not the excluded endpoint −π-\pi.

    Positive imaginary · x = 0, y > 0Quarter turn anticlockwise

    θ=π/2\theta=\pi/2

    Negative imaginary · x = 0, y < 0Quarter turn clockwise

    θ=−π/2\theta=-\pi/2

    Do not divide by x for a point on the imaginary axis. Its direction is read geometrically even though y/x is undefined.

  3. Use the basic angle and both coordinate signs

    Use this stage only when both coordinates are nonzero. Calculate the positive acute basic angle α in radians, then match the quadrant.

    α=tan⁡−1∣y/x∣\alpha=\tan^{-1}|y/x|

    0<α<π/20<\alpha<\pi/2

    I · x > 0, y > 0Use the basic angle

    θ=α\theta=\alpha

    II · x < 0, y > 0Subtract from π

    θ=π−α\theta=\pi-\alpha

    III · x < 0, y < 0Start at −π and add the basic angle

    θ=−π+α\theta=-\pi+\alpha

    IV · x > 0, y < 0Use the negative basic angle

    θ=−α\theta=-\alpha

    The ratio y/xy/x alone cannot distinguish opposite quadrants. Check both signs on the sketch and confirm −π<θ≤π-\pi<\theta\le\pi. Other arguments differ by 2kπ2k\pi for an integer k; select the representative in the stated interval.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

A negative real number uses the included endpoint

In Example 15(b), z=−3z=-3. Find its modulus and choose its principal argument.

Show reasoning

  1. 1. Locate the nonzero point

    The point (−3, 0) is on the negative real axis. It is not the origin, so it has a direction.

  2. 2. Separate distance from direction

    The squared coordinates sum to 9, so the distance is 3. The principal direction uses the included endpoint π.

    ∣−3∣=9=3,arg⁡(−3)=π.\begin{aligned}|-3|&=\sqrt9\\&=3,\\\arg(-3)&=\pi.\end{aligned}

  3. 3. Check the endpoint

    A clockwise half-turn gives −π, which reaches the same ray but is excluded from this principal interval. A zero imaginary part does not imply zero argument.

Continue with the lesson question

Equal negative coordinates do not give a first-quadrant angle

In Example 15(h), z=−1−iz=-1-i. Find its modulus and choose its principal argument.

Show reasoning

  1. 1. Use both signs before the quotient

    The point (−1, −1) lies in quadrant III. Its y/x ratio is positive, just as in quadrant I, so the quotient alone cannot select the direction.

    ∣−1−i∣=1+1=2.\begin{aligned}|-1-i|&=\sqrt{1+1}\\&=\sqrt2.\end{aligned}

  2. 2. Calculate the acute basic angle

    The absolute ratio gives the acute angle to the nearest horizontal ray, not the final principal argument.

    α=tan⁡−1∣−1−1∣=π4.\alpha=\tan^{-1}\left|\frac{-1}{-1}\right|=\frac\pi4.

  3. 3. Choose the clockwise principal direction

    Quadrant III requires a negative angle in the chosen interval. The positive turn 5π/4 reaches the same ray but is outside it.

    arg⁡(−1−i)=−π+π4=−3π4.\begin{aligned}\arg(-1-i)&=-\pi+\frac\pi4\\&=-\frac{3\pi}{4}.\end{aligned}
Continue with the lesson question

Compare with the eight Argand positions

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Eight Argand panels retain the source axis and quadrant cases: positive real 0, negative real pi, positive imaginary pi/2, negative imaginary -pi/2, quadrant I alpha, II pi-alpha, III -pi+alpha, IV -alpha. Blue arrows measure signed theta from the positive real axis; purple basic-angle arcs lie beside the nearest horizontal direction. Zero has no argument.
Eight Argand panels retain the source axis and quadrant cases: positive real 0, negative real pi, positive imaginary pi/2, negative imaginary -pi/2, quadrant I alpha, II pi-alpha, III -pi+alpha, IV -alpha. Blue arrows measure signed theta from the positive real axis; purple basic-angle arcs lie beside the nearest horizontal direction. Zero has no argument.
Eight Argand panels retain the source axis and quadrant cases: positive real 0, negative real pi, positive imaginary pi/2, negative imaginary -pi/2, quadrant I alpha, II pi-alpha, III -pi+alpha, IV -alpha. Blue arrows measure signed theta from the positive real axis; purple basic-angle arcs lie beside the nearest horizontal direction. Zero has no argument.
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