Clear Notion
Login

Concepts

Read through concept notes with worked explanations and diagrams, chapter by chapter.

Concept 01 / 10Complex Numbers and Quadratic Equations

Argand Plane, Modulus, Argument and the Polar and Euler Forms

Complex numbers as points and vectors, the modulus as distance, the principal argument in (-pi, pi] and the quadrant rule, the polar form r(cos t + i sin t), Euler's form re^(it), and why arguments add under multiplication.

Figure 1
Because a complex number x+iyx + iy is fixed by the ordered pair (x,y)(x, y), it can be drawn as the point (x,y)(x, y) in a plane, called the Argand plane (or complex plane), with the real axis horizontal and the imaginary axis vertical. It is equally useful to see zz as the vector from the origin to that point. Addition of complex numbers is then vector addition by the parallelogram law, and z2−z1z_{2} - z_{1} is the vector from z1z_{1} to z2z_{2}. Modulus: ∣z∣=x2+y2|z| = \sqrt{x^{2} + y^{2}} is the distance from the origin to zz. More usefully, ∣z1−z2∣|z_{1} - z_{2}| is the distance between the points z1z_{1} and z2z_{2}, and this single reading turns most locus problems into geometry: ∣z−z0∣=r|z - z_{0}| = r is a circle, ∣z−z1∣=∣z−z2∣|z - z_{1}| = |z - z_{2}| is the perpendicular bisector of the segment joining z1z_{1} and z2z_{2}, and so on. Argument: an argument of z≠0z \neq 0 is any angle θ\theta measured from the positive real axis to the vector zz, with anticlockwise positive. Arguments are determined only up to adding multiples of 2π2\pi. The principal argument, written Arg(z)\text{Arg}(z) or simply arg⁡z\arg z in JEE, is the unique value in (−π,π](-\pi, \pi]. The argument of 00 is undefined. Computing it with tan⁡−1yx\tan^{-1}\frac{y}{x} alone is the most common error, because tan⁡−1\tan^{-1} only returns angles in (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}) and cannot tell 1+i1 + i from −1−i-1 - i. The reliable method: find the acute angle α=tan⁡−1∣yx∣\alpha = \tan^{-1}|\frac{y}{x}|, then place it by quadrant, as shown in the figure. First quadrant (x>0,y>0x > 0, y > 0): θ=α\theta = \alpha. Second quadrant (x<0,y>0x < 0, y > 0): θ=π−α\theta = \pi - \alpha. Third quadrant (x<0,y<0x < 0, y < 0): θ=−(π−α)\theta = -(\pi - \alpha). Fourth quadrant (x>0,y<0x > 0, y < 0): θ=−α\theta = -\alpha. On the axes: positive real, 00; negative real, π\pi; positive imaginary, π2\frac{\pi}{2}; negative imaginary, −π2-\frac{\pi}{2}. For example, arg⁡(−1−i3)=−(π−π3)=−2π3\arg(-1 - i\sqrt{3}) = -(\pi - \frac{\pi}{3}) = -\frac{2\pi}{3}, not π3\frac{\pi}{3} as a calculator's tan⁡−13\tan^{-1}\sqrt{3} would suggest. Polar form: with r=∣z∣r = |z| and θ=arg⁡z\theta = \arg z, we have x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta, so z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta), often abbreviated r cis θr\,\text{cis}\,\theta. Euler's formula eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos\theta + i\sin\theta (which can be justified by comparing the power series of exe^{x}, cos⁡x\cos x and sin⁡x\sin x) gives the compact exponential form z=reiθz = re^{i\theta}. A famous special case is eiπ=−1e^{i\pi} = -1. Note that ∣eiθ∣=1|e^{i\theta}| = 1 for every real θ\theta, so eiθe^{i\theta} always lies on the unit circle. Why the polar form matters: multiplication becomes simple. r1eiθ1⋅r2eiθ2=r1r2ei(θ1+θ2)r_{1}e^{i\theta_{1}} \cdot r_{2}e^{i\theta_{2}} = r_{1}r_{2}e^{i(\theta_{1} + \theta_{2})}. So moduli multiply and arguments add: ∣z1z2∣=∣z1∣∣z2∣|z_{1}z_{2}| = |z_{1}||z_{2}| and arg⁡(z1z2)=arg⁡z1+arg⁡z2\arg(z_{1}z_{2}) = \arg z_{1} + \arg z_{2} (up to a multiple of 2π2\pi). Similarly ∣z1z2∣=∣z1∣∣z2∣|\frac{z_{1}}{z_{2}}| = \frac{|z_{1}|}{|z_{2}|} and arg⁡z1z2=arg⁡z1−arg⁡z2\arg\frac{z_{1}}{z_{2}} = \arg z_{1} - \arg z_{2}, and arg⁡zˉ=−arg⁡z\arg\bar{z} = -\arg z. The qualification 'up to 2π2\pi' matters for principal values. With z1=z2=−1z_{1} = z_{2} = -1, Arg(z1)+Arg(z2)=2π\text{Arg}(z_{1}) + \text{Arg}(z_{2}) = 2\pi, while Arg(z1z2)=Arg(1)=0\text{Arg}(z_{1}z_{2}) = \text{Arg}(1) = 0. The sum must be adjusted by ±2π\pm 2\pi to return to (−π,π](-\pi, \pi]. Key Takeaways: - ∣z1−z2∣|z_{1} - z_{2}| is the distance between the points z1z_{1} and z2z_{2}. - Principal argument lies in (−π,π](-\pi, \pi]; find the acute angle first, then use the quadrant. - arg⁡0\arg 0 is undefined; arg⁡\arg of a negative real number is π\pi. - z=r(cos⁡θ+isin⁡θ)=reiθz = r(\cos\theta + i\sin\theta) = re^{i\theta}; ∣eiθ∣=1|e^{i\theta}| = 1. - ∣z1z2∣=∣z1∣∣z2∣|z_{1}z_{2}| = |z_{1}||z_{2}| and arg⁡(z1z2)=arg⁡z1+arg⁡z2+2kπ\arg(z_{1}z_{2}) = \arg z_{1} + \arg z_{2} + 2k\pi, with kk chosen to return to the principal range. - arg⁡zˉ=−arg⁡z\arg\bar{z} = -\arg z (for zz not a negative real), and arg⁡(1z)=−arg⁡z\arg(\frac{1}{z}) = -\arg z. - Pitfall: using tan⁡−1yx\tan^{-1}\frac{y}{x} directly in the second or third quadrant. Brain-Teaser 1: Find the modulus and principal argument of z=1+i31−iz = \frac{1 + i\sqrt{3}}{1 - i} and hence z6z^{6}. Solution: Work in polar form rather than dividing. 1+i3=2eiπ/31 + i\sqrt{3} = 2e^{i\pi/3} and 1−i=2e−iπ/41 - i = \sqrt{2}e^{-i\pi/4}. So z=22ei(π/3+π/4)=2ei⋅7π/12z = \frac{2}{\sqrt{2}}e^{i(\pi/3 + \pi/4)} = \sqrt{2}e^{i \cdot 7\pi/12}. Thus ∣z∣=2|z| = \sqrt{2} and arg⁡z=7π12\arg z = \frac{7\pi}{12}, which is already in (−π,π](-\pi, \pi]. Then z6=(2)6ei⋅42π/12=8ei⋅7π/2z^{6} = (\sqrt{2})^{6}e^{i \cdot 42\pi/12} = 8e^{i \cdot 7\pi/2}. Since 7π2=4π−π2\frac{7\pi}{2} = 4\pi - \frac{\pi}{2}, this is 8e−iπ/2=−8i8e^{-i\pi/2} = -8i. Converting to polar form first makes both the quotient and the power almost mechanical. Brain-Teaser 2: Find the principal argument of z=1+cos⁡θ+isin⁡θz = 1 + \cos\theta + i\sin\theta for π<θ<2π\pi < \theta < 2\pi. Solution: Use half-angle identities: 1+cos⁡θ=2cos⁡2θ21 + \cos\theta = 2\cos^{2}\frac{\theta}{2} and sin⁡θ=2sin⁡θ2cos⁡θ2\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}. So z=2cos⁡θ2(cos⁡θ2+isin⁡θ2)z = 2\cos\frac{\theta}{2}(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}). It is tempting to read off ∣z∣=2cos⁡θ2|z| = 2\cos\frac{\theta}{2} and arg⁡z=θ2\arg z = \frac{\theta}{2}. But for π<θ<2π\pi < \theta < 2\pi, θ2∈(π2,π)\frac{\theta}{2} \in (\frac{\pi}{2}, \pi), so cos⁡θ2<0\cos\frac{\theta}{2} < 0, and a modulus cannot be negative. Absorb the minus sign: z=−2cos⁡θ2⋅(−1)eiθ/2=(−2cos⁡θ2)ei(θ/2−π)z = -2\cos\frac{\theta}{2} \cdot (-1) e^{i\theta/2} = (-2\cos\frac{\theta}{2})e^{i(\theta/2 - \pi)}. Now ∣z∣=−2cos⁡θ2>0|z| = -2\cos\frac{\theta}{2} > 0 and arg⁡z=θ2−π\arg z = \frac{\theta}{2} - \pi, which lies in (−π2,0)(-\frac{\pi}{2}, 0). A quick check: at θ=3π2\theta = \frac{3\pi}{2}, z=1−iz = 1 - i, with argument −π4=3π4−π-\frac{\pi}{4} = \frac{3\pi}{4} - \pi. The lesson is that r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta) is a valid polar form only when r>0r > 0. Brain-Teaser 3: If ∣z1∣=∣z2∣|z_{1}| = |z_{2}| and arg⁡z1+arg⁡z2=0\arg z_{1} + \arg z_{2} = 0, how are z1z_{1} and z2z_{2} related? What if instead arg⁡z1−arg⁡z2=π\arg z_{1} - \arg z_{2} = \pi? Solution: In the first case, write z1=reiθz_{1} = re^{i\theta}; then z2=re−iθ=zˉ1z_{2} = re^{-i\theta} = \bar{z}_{1}. They are conjugates: mirror images in the real axis. In the second case, z2=rei(θ−π)=−reiθ=−z1z_{2} = re^{i(\theta - \pi)} = -re^{i\theta} = -z_{1}. They are negatives of each other: diametrically opposite points on the same circle about the origin. Translating conditions on modulus and argument directly into the exponential form is faster than any coordinate computation.

Practice this concept

Free · no account required · instant solutions

Argand Plane, Modulus, Argument and the Polar and Euler Forms — DPP

4 questionsSee the solution the moment you answer

Practice