Express complex number $(1+i \sqrt 3 )^{27}(-2-2i)^8 $in exponential form

Question

Express complex number $$(1+i \sqrt 3 )^{27}(-2-2i)^8 $$in exponential form

While expressing it in exponential form i have to keep the argument between [-π,π].

Answer 1

$$1+i\sqrt{3}=2\left(\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}\right)=2\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)=2e^{\frac{\pi}{3}i}$$ $$-2-2i=-2\sqrt{2}\left(\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2}\right)=-2\sqrt{2}\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)=-2\sqrt{2}e^{\frac{\pi}{4}i}$$ therefore $$(1+i\sqrt{3})^{27}(-2-2i)^8=2^{27}e^{\frac{27\pi}{3}i}(-2\sqrt{2})^8e^{\frac{8\pi}{4}i}=2^{39}e^{11\pi i}=-2^{39}$$

Answer 2

Let's chop the the expression $(1+i\sqrt{3})^{27}(-2-2i)^8$ into $$z_1^{27}z_2^8$$ where $$z_1 = 1+i\sqrt{3}$$ and $$z_2 = -(2+2i) = e^{i\pi}(2+2i)$$ Let's start with $z_1$, we know that the modulus of $z_1 $ is $$\vert z_1 \vert = \sqrt{1^2 + \sqrt{3}^2} = 2$$ and the angle with the real axis would be $$\phi_1 = \arctan \frac{\sqrt{3}}{1} = \frac{\pi}{3}$$ So $$z_1 = \vert z_1 \vert e^{i \phi_1} = 2 e^{i\frac{\pi}{3}}$$

Let's jump to $z_2$, we know that the modulus of $z_2$ is $$\vert z_2 \vert = \sqrt{(-2)^2 + (-2)^2} = 2\sqrt{2}$$ and the angle with the real axis would be $$\phi_2 = \arctan \frac{\sqrt{-2}}{-2} = \frac{\pi}{4}$$ So $$z_2 = \vert z_2 \vert e^{i (\phi_2 + \pi)} = 2\sqrt{2} e^{i\frac{5\pi}{4}}$$

So $$ z_1^{27}z_2^8= ( 2 e^{i\frac{\pi}{3}})^{27} (2\sqrt{2} e^{i\frac{5\pi}{4}})^8 = 2^{27+8+4}e^{i(\frac{27\pi}{3} + 10\pi)}$$ that is $$ z_1^{27}z_2^8= 2^{39}e^{i9\pi}=2^{39}e^{i\pi}$$

Answer 3

Alternatively: $$\begin{align}(-2-2i)^8&=2^8((1+i)^2)^4=2^8(2i)^4=2^{12};\\ (1+i \sqrt 3 )^{27}&=((1+i\sqrt{3})^3)^9=\require{cancel}(1+\cancel{3\sqrt{3}i}-9-\cancel{3\sqrt{3}i})^9=-2^{27};\\ (1+i \sqrt 3 )^{27}(-2-2i)^8&=2^{12}\cdot (-2^{27})=-2^{39}.\end{align}$$