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tan⁻¹(y/x) gives the right argument of a complex number only half the time. For −1 − i it gives π/4, while the answer is −3π/4. The fix is a short set of quadrant rules, plus the habit of checking where the point actually is. This post gives the rules, works examples in all four quadrants and on the axes, and shows how to correct sums of arguments.
Where is this in the syllabus?
This is a topic NCERT has cut but JEE still asks. The NCERT Class 11 chapter Complex Numbers and Quadratic Equations (2026-27 reprint) keeps the heading "Argand Plane and Polar Representation" in section 4.5, but the text covers only the Argand plane, the modulus as distance and the conjugate as a reflection; the polar form and argument are not there. The JEE (Main) 2026 syllabus lists "modulus and argument (or amplitude) of a complex number", and the JEE (Advanced) 2026 brochure lists "polar representation, properties of modulus and principal argument". So if you are preparing for JEE, learn it from a reference book or the older NCERT edition. Check your year's syllabus.
What is the principal argument?
Write z = r(cos θ + i sin θ) with r = |z| > 0. Any θ that works is an argument; adding 2π gives another. The principal argument is the one in −π < θ ≤ π. Some books use 0 ≤ θ < 2π instead, so read the question. The argument of 0 is not defined.
What are the quadrant rules?
First find the acute reference angle α = tan⁻¹|y/x|. Then:
First quadrant (x > 0, y > 0): θ = α.
Second quadrant (x < 0, y > 0): θ = π − α.
Third quadrant (x < 0, y < 0): θ = −(π − α) = α − π.
Fourth quadrant (x > 0, y < 0): θ = −α.
On the axes: positive real axis 0, negative real axis π, positive imaginary axis π/2, negative imaginary axis −π/2.
Worked examples in each quadrant
z = 2 + 2√3 i. First quadrant. α = tan⁻¹(√3) = π/3, so arg z = π/3.
z = −√3 + i. Second quadrant. α = tan⁻¹(1/√3) = π/6, so arg z = π − π/6 = 5π/6.
z = −1 − √3 i. Third quadrant. α = tan⁻¹(√3) = π/3, so arg z = −(π − π/3) = −2π/3.
z = 3 − 3i. Fourth quadrant. α = π/4, so arg z = −π/4.
z = −4 and z = −2i. On the axes: π and −π/2.
Each of these can be checked numerically: Python's cmath.phase returns the principal argument, and gives 5π/6, −2π/3 and −π/4 for the middle three.
Worked example: simplify first
Find arg z for z = (1 + i)/(1 − i).
Multiply top and bottom by (1 + i): z = (1 + i)²/2 = 2i/2 = i. So arg z = π/2. You can also get this from arguments: arg(1 + i) − arg(1 − i) = π/4 − (−π/4) = π/2. Both routes agree, and here no correction is needed.
Why does arg(z1z2) sometimes need a correction?
The rule arg(z1z2) = arg z1 + arg z2 holds up to a multiple of 2π. If the sum leaves (−π, π], bring it back.
Take z1 = −1 + i and z2 = −1 + √3 i.
arg z1 = 3π/4 and arg z2 = 2π/3. The sum is 17π/12, which is more than π.
Subtract 2π: arg(z1z2) = 17π/12 − 2π = −7π/12.
Check by multiplying: z1z2 = (1 − √3) − (1 + √3)i, roughly −0.73 − 2.73i. That is in the third quadrant, so a negative argument between −π and −π/2 is right, and −7π/12 is about −105°.
The same happens with powers. For z = −1 + i, 3 arg z = 9π/4, so the principal argument of z³ is 9π/4 − 2π = π/4. Check: z³ = 2 + 2i.
Common mistakes and how to avoid them
Using tan⁻¹(y/x) with signs. For z = −1 − i, tan⁻¹(1) = π/4, the wrong quadrant. Use |y/x| and place the angle yourself.
Writing 4π/3 instead of −2π/3. Both are arguments, but 4π/3 is outside (−π, π].
Giving −π for the negative real axis. The interval includes π and excludes −π, so the answer is π.
Adding arguments without checking the range. Always bring the sum back into (−π, π].
For the geometry that uses arguments, such as loci and rotations, see complex numbers as an Argand-plane tool. For solving equations by comparing parts, see solving equations by putting z = x + iy.
A practice routine
Day 1: Twelve numbers, three per quadrant, with standard angles. Sketch each before you calculate.
Day 2: Four quotients and products, simplified both ways.
Day 3: Arguments of z², z³ and 1/z for three numbers, corrected to the principal range.
Day 4: Redo Day 1 in the 0 to 2π convention and compare.
If you can see where the point lies but cannot picture the rotation, Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Tonight, find the principal arguments of −2 + 2i, −5 and (1 + √3 i)², and check each against a sketch.
