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

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:

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 n^th 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 n^th 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