Last modified on February 22nd, 2024

chapter outline

 

The Polar Form of a Complex Number

The polar form of a complex number z = a + ib represents the complex number in terms of r and θ, where r = length of the point from the origin, and θ = angle made with the x-axis in an argand plane. 

Plotting the number z in a cartesian plane, we get

The Polar Form of a Complex Number

Here, (a, b) are the rectangular coordinates, and the x, y axes represent the real and imaginary parts, respectively.

Now, using the Pythagorean theorem and the basic trigonometric ratio, we get r2 = a2 + b2 and Cosθ = ${\dfrac{a}{r}}$ and Sinθ = ${\dfrac{b}{r}}$.

⇒ a = rcosθ, b = rsinθ

Substituting the values of a and b in z = a + ib, we get

z = rcosθ + irsinθ

Thus, the equation of any complex number z = a + ib in the polar form is

z = r(cosθ + isinθ), here 

  • the polar coordinates are (r, θ),
  • r = modulus or the absolute value of z = ${\sqrt{a^{2}+b^{2}}}$, 
  • a = rcosθ, 
  • b = rsinθ, and
  • θ =  the argument of z = ${\tan ^{-1}\left( \dfrac{b}{a}\right)}$ for a > 0
  • θ = ${\tan ^{-1}\left( \dfrac{b}{a}\right)}$ + π

Or, θ = ${\tan ^{-1}\left( \dfrac{b}{a}\right)}$ + 180° for a < 0

Converting From Rectangular to Polar Form

Now, let us convert a complex number z = -3 + 3i in the polar form. Since the real part is -3, z lies in the second quadrant, as shown:

Complex Number Into Polar Form

Here, we calculate the modulus r and the argument θ of z

r = |z| = ${\sqrt{\left( -3\right) ^{2}+\left( 3\right) ^{2}}}$ = ${\sqrt{18}}$ ≈ 4.24

θ = ${\tan ^{-1}\left( \dfrac{3}{-3}\right)}$ = -${\tan ^{-1}\left( \dfrac{3}{3}\right)}$.

Since z lies in the second quadrant, arg(z) = θ = -${\tan ^{-1}\left( \dfrac{3}{3}\right)}$ + 180°

= -45° + 180° = 135°

Thus, the polar form of the complex number z = -3 + 3i is 4.24(cos135° + isin135°).

Simplifying

Adding

Let 5 + 3i and 2(cos60° + isin60°) be two complex numbers, one in the standard (rectangular) form and another in the polar form.

On adding, first, we convert 2(cos60° + isin60°) in the polar form into the standard form.

z = 2(cos60° + isin60°) = a + ib, here a = 2cos60° = 0.5 and b = 2sin60° = ${\sqrt{3}}$

⇒ 2(cos60° + isin60°) = 0.5 + ${i\sqrt{3}}$

Adding both the given complex numbers, we get (5 + 3i) + (0.5 + ${i\sqrt{3}}$) = (5 + 0.5) + (3 + ${\sqrt{3}}$)i.

By substituting the value of ${\sqrt{3}}$ ≈ 1.732, the sum is 5.5 + 4.732i

Now, we convert the sum in the polar form by finding the modulus and argument of the number.

r = |z| = ${\sqrt{\left( 5.5\right) ^{2}+\left( 4.732\right) ^{2}}}$ ≈ ${\sqrt{52.64}}$, and

θ = ${\tan ^{-1}\left( \dfrac{4.732}{5.5}\right)}$ ≈ 0.71

Thus, the required sum of 5 + 3i and 2(cos60° + sin60°) is ${\sqrt{52.64}}$(cos0.71° + sin0.71°)

Multiplying

On multiplying z = 4(cos90° + isin90°) by w = 2(cos240° + isin240°), the product obtained is:

zw = 4(cos90° + isin90°) ✕ 2(cos240° + isin240°)

= 4 ✕ 2 [(cos90°cos240° – sin90°sin240°) + i(sin90°cos240° + cos90°sin240°)]

= 8[cos(90° + 240°) + isin(90° + 240°)]

= 8(cos330° + isin330°)

Thus, for z1 = r1(cosθ1 + isinθ1) and z2 = r2(cosθ2 + isinθ2), the product will be

z1z2 = r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]

Dividing

On dividing z = 6(cos150° + isin150°) by w = 12(cos90° + isin90°), the quotient is:

${\dfrac{z}{w}}$

= ${\dfrac{6\left( \cos 150^{\circ }+i\sin 150^{\circ }\right) }{12\left( \cos 90^{\circ }+i\sin 90^{\circ }\right) }\times \dfrac{\left( \cos 90^{\circ }-iSin90^{\circ }\right) }{\left( \cos 90^{\circ }-i\sin 90^{\circ }\right) }}$

= ${\dfrac{1}{2}\dfrac{\left( \cos 150^{\circ }cos90^{\circ }+\sin 150^{\circ }sin90^{\circ }\right) +i\left( \sin 150^{\circ }cos90^{\circ }-\cos150^{\circ }sin90^{\circ } \right) }{\cos ^{2}90^{\circ }-i^{2}\sin ^{2}90^{\circ }}}$

= ${\dfrac{1}{2}\dfrac{\cos \left( 150^{\circ }-90^{\circ }\right) +i\sin \left( 150^{\circ }-90^{\circ }\right) }{\cos ^{2}90^{\circ }+\sin ^{2}90^{\circ }}}$

= ${\dfrac{1}{2}\left( \cos 60^{\circ }+i\sin 60^{\circ }\right)}$

Thus, for z1 = r1(cosθ1 + isinθ1) and z2 = r2(cosθ2 + isinθ2), the quotient will be

${\dfrac{z_{1}}{z_{2}}=\dfrac{r_{1}\left( \cos \theta _{1}+i\sin \theta _{1}\right) }{r_{2}\left( \cos \theta _{2}+i\sin \theta _{2}\right) }}$

⇒ ${\dfrac{z_{1}}{z_{2}}=\dfrac{r_{1}}{r_{2}}\left[ \cos \left( \theta _{1}-\theta _{2}\right) +i\sin \left( \theta _{1}-\theta _{2}\right) \right]}$

${\dfrac{z_{1}}{z_{2}}=r\left( \cos \theta +i\sin \theta \right)}$, here r = ${\dfrac{r_{1}}{r_{2}}}$, and θ = θ1 – θ2.

Finding Powers

The powers of a complex number can easily be found by De Moivre’s theorem. It states that for any natural number n, zn is calculated by increasing the modulus to the nth power and multiplying the argument by n.

If z = r(cosθ + isinθ), then it is mathematically written as

zn = rn[cos(nθ) + isin(nθ)], where n is any natural number.

Let us evaluate an expression z3 = (5 + 5i)3 using De Moivre’s theorem.

First, we find the modulus value r and argument angle θ.

r = ${\sqrt{5^{2}+5^{2}}}$ = ${\sqrt{25+25}}$ = ${\sqrt{50}}$

θ = ${\tan ^{-1}\dfrac{5}{5}}$ = 45°

Using De Moivre’s theorem, we get z3 = (5 + 5i)3 = r5[cos(5θ) + isin(5θ)]

= ${\left( \sqrt{50}\right) ^{3}\left[ \cos \left( 5\times 45^{\circ }\right) +i\sin \left( 5\times 45^{\circ }\right) \right]}$

= ${250\sqrt{2}\left( \cos 225^{\circ }+i\sin 225^{\circ }\right)}$

Finding Roots

To find the nth root of a complex number in the polar form, we use the formula:

${z^{\dfrac{1}{n}}=r^{\dfrac{1}{n}}\left[ \cos \left( \dfrac{\theta }{n}+\dfrac{2k\pi }{n}\right) +i\sin \left( \dfrac{\theta }{n}+\dfrac{2k\pi }{n}\right) \right]}$, where k = 0, 1, 2, …, n – 1.

Let us evaluate the cube root of an expression z = ${27\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4}\right)}$

${z^{\dfrac{1}{3}}=27^{\dfrac{1}{3}}\left[ \cos \left( \dfrac{\dfrac{\pi }{4}}{3}+\dfrac{2k\pi }{3}\right) +i\sin \left( \dfrac{\dfrac{\pi }{4}}{3}+\dfrac{2k\pi }{3}\right) \right]}$

= ${3\left[ \cos \left( \dfrac{\pi }{12}+\dfrac{2k\pi }{3}\right) +i\sin \left( \dfrac{\pi }{12}+\dfrac{2k\pi }{3}\right) \right]}$

There are three roots: for k = 0, 1, 2.

When k = 0, we get 

${z^{\dfrac{1}{3}}=3\left( \cos \dfrac{\pi }{12}+i\sin \dfrac{\pi }{12}\right)}$

When k = 1, we get

${z^{\dfrac{1}{3}}=3\left[ \cos \left( \dfrac{\pi }{12}+\dfrac{2\pi }{3}\right) +i\sin \left( \dfrac{\pi }{12}+\dfrac{2\pi }{3}\right) \right]}$

⇒ ${z^{\dfrac{1}{3}}=3\left( \cos \dfrac{3\pi }{4}+i\sin \dfrac{3\pi }{4}\right)}$

When k = 2, we get

${z^{\dfrac{1}{3}}=3\left[ \cos \left( \dfrac{\pi }{12}+\dfrac{4\pi }{3}\right) +i\sin \left( \dfrac{\pi }{12}+\dfrac{4\pi }{3}\right) \right]}$

⇒ ${z^{\dfrac{1}{3}}=3\left( \cos \dfrac{17\pi }{12}+i\sin \dfrac{17\pi }{12}\right)}$

Solved Examples

Plot the complex number z = -2 + 4i in the argand plane.

Solution:

Find the polar form of the complex number z = 8 + 8i.

Solution:

r = ${\sqrt{\left( 8\right) ^{2}+\left( 8\right) ^{2}}}$ = ${\sqrt{64+64}}$
= ${\sqrt{128}}$ = ${8\sqrt{2}}$
θ = ${\tan ^{-1}\left( \dfrac{8}{8}\right)}$ = ${\tan ^{-1}1}$ ≈ 45°
Thus, the polar form of the given complex number z = 8 + 8i is ${8\sqrt{2}}$(cos45° + isin45°)

Last modified on February 22nd, 2024

Comments are closed.