Find the polar form of complex number Z = 3 + 3i.

CONCEPT:

Modulus of complex number:

Let the point P represent the nonzero complex number z = x + iy.

Here \(r = \sqrt {{x^2} + {y^2}} = \left| z \right|\) is called modulus of given complex number.

Argument of complex number:

For any complex number z ≠ 0, there corresponds only one value of θ in 0 ≤ θ < 2π. However, any other interval of length 2π, for example – π < θ ≤ π, can be such an interval. We shall take the value of θ such that – π < θ ≤ π, called principal argument of z and is denoted by arg (z).

The argument of Z is measured from positive x-axis only.

Let z = r (cosθ + i sinθ) is polar form of any complex number then following ways are used while writing θ for different quadrants –

For first quadrant \(\left( {0 < {\rm{\theta }} \le \frac{{\rm{\pi }}}{2}} \right),\;\theta = {\tan ^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\)

For second quadrant \(\left( {\frac{{\rm{\pi }}}{2} < {\rm{\theta }} \le {\rm{\pi }}} \right),\;\theta = {\rm{\pi }} - {\tan ^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\)

For third quadrant \(\left( {{\rm{\pi }} < {\rm{\theta }} \le \frac{{3{\rm{\pi }}}}{2}} \right),\;{\rm{\theta }} = - {\rm{\pi }} + {\tan ^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\)

For fourth quadrant \(\left( {\frac{{3{\rm{\pi }}}}{2} < {\rm{\theta }} \le 2{\rm{\pi }}} \right),\;\theta = - {\rm{\;ta}}{{\rm{n}}^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\)

CALCULATION:

Given complex number Z = 3 + 3i.

 Let rcosθ = 3 and rsinθ = 3

By squaring and adding, we get

\({{\rm{r}}^2}\left( {{\rm{co}}{{\rm{s}}^2}{\rm{\theta }} + {\rm{si}}{{\rm{n}}^2}{\rm{\theta }}} \right){\rm{\;}} = {\rm{\;}}18\)

∴ r = 3√2

As we have taken rcosθ = 3

\(\Rightarrow cos\theta = \frac{3}{r} = \frac{3}{{3\sqrt 2 }} = \frac{1}{{\sqrt 2 }}\)

⇒ θ = 45° (Since it is in first quadrant, there won’t be any changes in θ value)

So, on comparing with z = r (cosθ + i sinθ), we can write as \(3\sqrt 2 \left( {\cos 45^\circ + i\;sin\;45^\circ } \right)\).