Question

express the complex number 3 vroot 2 (-1+i) into polar form​

Answer 1

$3 \sqrt{2}( - 1 + i) \\ \\ = - 3 \sqrt{2} + 3 \sqrt{2}i$

Now we know that polar form of complex number is :-

$r( \cos(a) + i \sin(a))$

Now,

on comparing we get :-

$r \cos(a) = - 3 \sqrt{2} \\ \\ {r}^{2} { \cos}^{2}a = 18 \\ \\ \\ r \sin(a) = 3 \sqrt{2} \\ \\ {r}^{2} { \sin}^{2}a = 18 \\ \\ \\ {r}^{2}( { \sin(a) }^{2} + { \cos(a) }^{2}) = 18 + 18 = 36 \\ \\ \\r = \sqrt{36} = 6$

Now we also know that:-

$\tan(a) = \frac{y}{x} \\ \\ \\ \tan(a) = \frac{3 \sqrt{2} }{ - 3\sqrt{2}} = \frac{1}{ - 1}$

It means it lies in 2nd Quadrant

so,

$\tan(a) = - \tan( \frac{\pi}{4}) \\ \\ \tan(a) = \tan( \frac{3\pi}{4} ) \\ \\ \\a = \frac{3\pi}{4}$

so,

Polar form of the complex Number is :-

$6( \cos( \frac{3\pi}{4} + i \sin( \frac{3\pi}{4})))$

Answer 2

$\begin{gathered}3 \sqrt{2}( - 1 + i) \\ \\ = - 3 \sqrt{2} + 3 \sqrt{2}i\end{gathered} </p><p>$

Now we know that polar form of complex number is :-

$r( \cos(a) + i \sin(a))$

Now,

on comparing we get :-

$\begin{gathered}r \cos(a) = - 3 \sqrt{2} \\ \\ {r}^{2} { \cos}^{2}a = 18 \\ \\ \\ r \sin(a) = 3 \sqrt{2} \\ \\ {r}^{2} { \sin}^{2}a = 18 \\ \\ \\ {r}^{2}( { \sin(a) }^{2} + { \cos(a) }^{2}) = 18 + 18 = 36 \\ \\ \\r = \sqrt{36} = 6\end{gathered}$

Now we also know that:-

$\begin{gathered}\tan(a) = \frac{y}{x} \\ \\ \\ \tan(a) = \frac{3 \sqrt{2} }{ - 3\sqrt{2}} = \frac{1}{ - 1} \\ \\\end{gathered}$

It means it lies in 2nd Quadrant

so,

$\begin{gathered}\tan(a) = - \tan( \frac{\pi}{4}) \\ \\ \tan(a) = \tan( \frac{3\pi}{4} ) \\ \\ \\a = \frac{3\pi}{4}\end{gathered}$

so,

Polar form of the complex Number is :-

$6( \cos( \frac{3\pi}{4} + i \sin( \frac{3\pi}{4})))$