Write the complex number -2-2i in its polar form
A) $2\left( \cos \left( \dfrac{\pi }{4} \right)+i\sin \left( \dfrac{\pi }{4} \right) \right)$
B) -$2\left( \cos \left( \dfrac{\pi }{4} \right)-i\sin \left( \dfrac{\pi }{4} \right) \right)$
C) $2\sqrt{2}\left( \cos \left( \dfrac{3\pi }{4} \right)+i\sin \left( \dfrac{3\pi }{4} \right) \right)$
D) $2\sqrt{2}\left( \cos \left( \dfrac{7\pi }{4} \right)+i\sin \left( \dfrac{7\pi }{4} \right) \right)$
E) $2\sqrt{2}\left( \cos \left( \dfrac{5\pi }{4} \right)+i\sin \left( \dfrac{5\pi }{4} \right) \right)$

Hint: In this question we have to find the polar form of a given complex number. Therefore, we should try to understand the method to obtain the polar form by finding the magnitude and argument of the given number and then we can use the Euler equation to write the resulting form in terms of sine and cosine of the angle.
Complete step-by-step answer:
In this question, we want to obtain the polar form of the complex number -2-2i. Therefore, we will use the following theorem to obtain the polar form of a complex number x+iy.
$\text{Polar form of the number }x+iy=r{{e}^{i\theta }}$
Where the parameters r and $\theta $ are given by
$r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ and $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$
(-2 - 2i) compared to the general complex number\[x+iy\], in this case, we have x = -2 and y = -2. Therefore,
$r=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -2 \right)}^{2}}}=2\sqrt{2}$ and $\theta ={{\tan }^{-1}}\left( \dfrac{-2}{-2} \right)={{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}+n\pi ..............(1.1)$
As in tangent of an angle and that of its sum with a multiple of $\pi $ is the same. However, we notice that we can express the argument between 0 and $2\pi $ only as after that the angles repeat. As here both x and y are negative the point (x,y) lies in the third quadrant, therefore, the argument should be between $\pi \text{ and }\dfrac{3\pi }{2}$. Comparing with equation (1.1), we get
$\theta =\dfrac{\pi }{4}+\pi =\dfrac{5\pi }{4}.................(1.2)$

Again, we note that the Euler’s equation states that
${{e}^{i\theta }}=\cos (\theta )+i\sin (\theta ).....................(1.3)$
Therefore, from equation (1.1), (1.2) and (1.3), we get
$-2-2i=2\sqrt{2}{{e}^{i\dfrac{5\pi }{4}}}=2\sqrt{2}\left( \cos \left( \dfrac{5\pi }{4} \right)+i\sin \left( \dfrac{5\pi }{4} \right) \right)$
Which matches option (E). Therefore, option(E) is the correct answer to this question.
Note: We should be careful to include the addition of $\pi $ in equation (1.2), because in the interval $\left( 0,2\pi \right)$, two angles may have the same tangent value, in that case we should look at the coordinates and infer the quadrant in which the point lies.