West Bengal WbchseTextbookSubjectiveImportance★★★★★est
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✓ Free question
Concept understanding — Modulus and Argument of a Complex Number
For a complex number z=a+ib represented by the point P(a,b) in the Argand plane, the modulus ∣z∣=r=a2+b2 is the straight-line distance OP from the origin to P (found via Pythagoras on the right triangle formed by the point and the two axes), and the argument θ=arg(z) is the angle that OP makes with the positive real axis, satisfying cosθ=a/r, sinθ=b/r, and tanθ=b/a when a=0. Because the inverse tangent function alone only ever returns an angle in (−π/2,π/2), it cannot on its own distinguish a point in quadrant I from one in quadrant III (or quadrant II from quadrant IV), so finding the argument in the standard range 0≤θ<2π requires checking which quadrant (or axis) the point (a,b) actually lies in and adding the appropriate correction: no correction in quadrant I, +π in quadrants II and III, and +2π in quadrant IV. The modulus obeys clean multiplicative rules — ∣z1z2∣=∣z1∣∣z2∣, z2z1=∣z2∣∣z1∣, and zzˉ=∣z∣2 — together with the triangle inequality ∣z1+z2∣≤∣z1∣+∣z2∣, while the argument obeys additive rules — arg(z1z2)=argz1+argz2 and arg(z1/z2)=argz1−argz2 — which together are the algebraic seeds of De Moivre's theorem.
Compute ∣z∣; both parts negative means Quadrant III.
✓Final answer
∣z∣=2, arg(z)=−65π.
For z=−3−i: ∣z∣=(3)2+12=3+1=2. Since a=−3<0 and b=−1<0, z lies in Quadrant III. Reference angle: tan−1(31)=6π. Using the principal range (−π,π], the Quadrant-III argument is θ=−(π−6π)=−65π. Check: 2(cos(−65π)+isin(−65π))=2(−23−i21)=−3−i, matching z.
✓Final answer
∣z∣=2, arg(z)=−65π.
Compute the modulus; identify Quadrant III from both parts negative; find the reference angle; apply the Quadrant-III correction for the (−π,π] convention.
Reporting the argument as the positive value 65π instead of the correct negative value under the (−π,π] convention
Using tan−1(b/a) directly, which for two negative numbers returns a Quadrant-I-looking positive reference angle only, not the true quadrant
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
West Bengal HS First Year (WBCHSE Class XI) Annual Examination 2018Set ANNUAL2 marks
Q.Find the principal amplitude of (-1-i).
›Reveal solutionSolution
−1−i lies in the third quadrant with reference angle π/4; the principal argument (in (−π,π]) is −3π/4.
For z=−1−i, both the real part (−1) and imaginary part (−1) are negative, so z lies in the third quadrant.
The reference angle is tan−1(∣−1∣∣−1∣)=tan−1(1)=4π.
For the principal value convention (argument in (−π,π]), a point in the third quadrant has argument −(π−4π)=−43π (measuring the shorter, negative/clockwise rotation from the positive real axis).