The correct answer is: $2\sqrt{2}\big(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\big)=2\sqrt{2}e^{i\pi/4}$
Explanation
We convert to polar form by finding the modulus and argument.
Steps:
- Modulus: $$r=\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}$$
- Argument (principal value): $$\theta=\arctan\frac{2}{2}=\arctan1=\frac{\pi}{4}$$
- Final polar form: $$2+2i=2\sqrt{2}\big(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\big)=2\sqrt{2}e^{i\pi/4}$$
Therefore, the polar representation is $2\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$ (or $2\sqrt{2}e^{i\pi/4}$).