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The complex number z = x + iy so

x = rcos$\Theta$ ; y = rsin $\Theta$ and tan$\Theta$ = $\left ( \frac{y}{x} \right )$

∴ z = r(rcos$\Theta$ + i sin $\Theta$) is the polar form (or the trigonometric form) of a complex number.

The polar form of the complex number |z|= $\sqrt{x^2 +y^2}$

|z| is the modulus of z and $\Theta$ is called the argument (or amplitude) of z which is denoted by arg z.

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

1 = rcos$\Theta$

Squaring both side

1 = $r^2$ $cos^{2}\Theta$ -----(1)

√3 = r sin$\Theta$

Squaring both side

3 = $r^2$ $sin^{2}\Theta$ -----(2)

Add equation (1) and (2)

1 + 3 = $r^2$ $cos^{2}\Theta$ + $r^2$ $sin^{2}\Theta$

4 = $r^2$ ($cos^{2}\Theta$ + $sin^{2}\Theta$ )

4 = $r^2$

∴ r = 2

So, 1 = 2 cos$\Theta$ ⇒ cos$\Theta$ = 1/2

⇒ $\Theta$ = $\frac{\prod }{3}$

√3 = 2 sin$\Theta$ ⇒ sin$\Theta$ = $\frac{\sqrt{3}}{2}$

⇒ $\Theta$ = $\frac{\prod }{3}$

Thus, required polar form = z = $\left ( 2cos\frac{\prod }{3} + sin\frac{\prod }{3} \right )$

z = ${\frac{-16}{1 + i\sqrt{3}}}$

= ${\frac{-16(1-i\sqrt{3})}{(1+i\sqrt{3})(1 -i\sqrt{3})}}$

= ${\frac{-16+ 16i\sqrt{3}}{(1+ 3)}}$

z =$\frac{-16+ 16i\sqrt{3}}{4}$

z = ${\frac{-16}{4}}$ + $\frac{16}{4}i\sqrt{3}$

z = - 4 + 4i$\sqrt{3}$

From the above we can say that,

-4 = r cos$\Theta$ and 4$\sqrt{3}$ = r sin$\Theta$

Squaring and then add we get,

16 + 48 = ${r^{2}(cos^{2}\Theta + sin^{2}\Theta)}$

64 =$r^{2}$

∴ r = 8

So - 4 = 8 cos$\Theta$ 4$\sqrt{3}$ = 8sin$\Theta$

$\frac{-1}{2}$ = cos$\Theta$ ${\frac{\sqrt{3}}{2}}$ = sin$\Theta$

Since cos $\Theta$ is negative,

∴ $\Theta$ = ${\prod}$ - $\frac{2\prod }{3}$ = ${\frac{2\prod }{3}}$

So Polar form is 8${\left ( cos\frac{2\prod }{3} + i sin\frac{2\prod }{3} \right )}$

From Complex numbers Polar form to Home

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