Question

Represent the 4+4√3i complex number in the polar form​

Answer 1

Answer:

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Answer 2

Answer:

Here we can find the polar form for the given complex number $4+4\sqrt{3}i$.

Final Answer is: $8(cos\frac{\pi }{3} +isin\frac{\pi }{3})$

Step-by-step explanation:

from the given question,

     complex number $Z=4+4\sqrt{3}i$           ----------------(1)

Polar form of the complex number is,

                                  $Z=r(cos\theta+isin\theta)$  ----------------(2)    

Now equate equation(1) and equation(2). Then we get the following equation.

                       $4+4\sqrt{3}i = r(cos\theta+i sin\theta)$  

Here we should find the value of $r$ and $\theta$.

From the above equation we can get the following values. That is,

                              $r cos\theta=4$    

                                $cos\theta=\frac{4}{r}$         -----------------(3)

                              $r sin\theta=4\sqrt{3}$  

                                $sin\theta=\frac{4\sqrt{3}}{r}$     -----------------(4)

For find the $\theta$ value, we should divide equation(3) by equation(4)

                                 $\frac{sin\theta}{cos\theta} = \frac{(4\sqrt{3}/r )}{4/r}$

It can be written as, $tan\theta=\frac{4\sqrt{3} }{r} * \frac{r}{4}$    (Trigonometric Formula:$tan \theta=\frac{sin\theta}{cos\theta}$)

here we cancel $r$ and $4$ in the numerator and denominator. Then we get the following equation.

                                 $tan\theta=\sqrt{3}$

Using Trigonometric table,

                                 $tan\theta=\sqrt{3} = tan \frac{\pi }{3}$

So that,                theta  $\theta=\frac{\pi }{3}$

Next we should find the value of $r$. For that we squared and add equation(3) and equation(4).

                    $sin^{2}\theta+ cos^{2}\theta = (\frac{4\sqrt{3} }{r} )^{2} + (\frac{4}{r} )^{2}$

                                       $1 = \frac{16}{r ^{2}}(\sqrt{3} )^2 + \frac{16}{r^{2}}$          ($sin^{2}\theta+ cos^{2}\theta = 1$)

                                       $1 = \frac{16}{r ^{2}}({3} ) + \frac{16}{r^{2}}$               ($\sqrt{3} ^2=3$)

                                       $1 = \frac{48}{r ^{2}} + \frac{16}{r^{2}}$                    (Add the numerator)

                                       $1 = \frac{64}{r ^{2}}$

                                      $r ^{2}= 64$

                                        $r=\sqrt{64}$

                        value of  $r=8$

Now we should apply the value of $r$ and $\theta$ in equation(2)

                                       $Z=r(cos\theta+isin\theta)$

Polar form of the complex number

                                       $Z=8(cos\frac{\pi }{3} + i sin\frac{\pi }{3} )$