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Mathematics · Ch 4 — Complex Numbers and Quadratic Equations

Polar Representation of a Complex Number

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Polar Representation of a Complex Number

Polar Representation of a Complex Number

From the right triangle used to define the modulus and argument (§4), the real and imaginary parts of z=a+ibz=a+ib can themselves be written in terms of r=∣z∣r=|z| and θ=arg⁡(z)\theta=\arg(z):

a=rcos⁡θ,b=rsin⁡θ.a=r\cos\theta, \qquad b=r\sin\theta.

Substituting these into z=a+ibz=a+ib gives the polar form:

z=rcos⁡θ+i rsin⁡θ=r(cos⁡θ+isin⁡θ).z=r\cos\theta+i\,r\sin\theta=r(\cos\theta+i\sin\theta).

Here r=∣z∣≥0r=|z|\ge0 is the modulus and θ=arg⁡(z)\theta=\arg(z) is the argument (usually taken as the principal value in (−π,π](-\pi,\pi] unless stated otherwise). The ordered pair (r,θ)(r,\theta) is called the polar coordinates of the point PP representing zz, and by tradition the origin OO is called the pole in this description.

Converting Cartesian →\to polar means computing r=a2+b2r=\sqrt{a^2+b^2} and finding θ\theta from the quadrant-corrected inverse tangent (§4), then writing z=r(cos⁡θ+isin⁡θ)z=r(\cos\theta+i\sin\theta).

z=1+i3z=1+i\sqrt3: r=1+3=2r=\sqrt{1+3}=2; since a=1>0,b=3>0a=1>0,b=\sqrt3>0 (Quadrant I), θ=tan⁡−1(3/1)=π/3\theta=\tan^{-1}(\sqrt3/1)=\pi/3. So z=2(cos⁡π3+isin⁡π3)z=2\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right).

Converting polar →\to Cartesian means simply evaluating cos⁡θ\cos\theta and sin⁡θ\sin\theta for the given angle and multiplying through by rr.

For r=2, θ=π/3r=2,\ \theta=\pi/3: cos⁡(π/3)=12\cos(\pi/3)=\tfrac12, sin⁡(π/3)=32\sin(\pi/3)=\tfrac{\sqrt3}2, so z=2(12+i32)=1+3 iz=2\left(\tfrac12+i\tfrac{\sqrt3}2\right)=1+\sqrt3\,i. …

Figure Fig 2Polar coordinates (r, theta) of the point representing z

What this figure shows. The same point PP representing z=a+ibz=a+ib from the Argand-plane figure is redrawn with the segment OPOP, of length r=∣z∣r=|z|, making an angle θ\theta with the positive real axis, measured anticlockwise when θ>0\theta>0. Dashed perpendiculars from PP to each axis reproduce the right triangle with legs a=rcos⁡θa=r\cos\theta (along the real axis) and b=rsin⁡θb=r\sin\theta (along the imaginary axis), which is exactly the substitution used to derive the polar form z=r(cos⁡θ+isin⁡θ)z=r(\cos\theta+i\sin\theta) from the Cartesian form z=a+ibz=a+ib. The figure is annotated to show that (r,θ)(r,\theta) and (a,b)(a,b) describe the very same point PP, just in two diffe …