Converting a complex number to polar form

Question

I'm studying for a test and doing old homework exercises and struggling with this one (excuse me for not being allowed to use html in here):

Calculate the polar coordinates of the following complex number: $$(1+i) (\sqrt{2}i) (-1+i) (-\sqrt{2}i) (-1-i)$$

My first idea was to multiply each of them with each other and then calculate the coordinates from the resulting complex number. My second one is to calculate the polar coordinates of each of them and then multiply them. With both attempts I fail to get to the suggested solution of:

π/4 * √(2)^5

I would be thankful for help

Answer 1

Use, when $\text{z}\in\mathbb{C}$:

$$\text{z}\cdot\overline{\text{z}}=\left|\text{z}\right|^2=\Re^2[\text{z}]+\Im^2[\text{z}]$$

So, for:

$$\text{q}=(1+i)\cdot(\sqrt{2}i)\cdot(-1+i)\cdot(-\sqrt{2}i)\cdot(-1-i)$$

See that:

• $$(-1+i)\cdot(-1-i)=(-1+i)\cdot\overline{-1+i}=(-1)^2+1^2=1+1=2$$
• $$(\sqrt{2}i)\cdot(-\sqrt{2}i)=(\sqrt{2}i)\cdot\overline{\sqrt{2}i}=(0)^2+(\sqrt{2})^2=0+2=2$$

Now, we get:

$$\text{q}=2\cdot2\cdot(1+i)=4\cdot(1+i)=4+4i$$

Now, in the polar form:

$$\text{q}=|4+4i|e^{\left(\arg(4+4i)+2\pi k\right)i}=\sqrt{4^2+4^2}e^{\left(\arctan(1)+2\pi k\right)i}=4\sqrt{2}e^{\left(\frac{\pi}{4}+2\pi k\right)i}$$

Where $k\in\mathbb{Z}$

Answer 2

You have an unusual notation for the result. All factors have magnitude $\sqrt{2}$ and the phase angles are (in this order): $$\frac{\pi}{4}, \frac{2\pi}{4}, \frac{3 \pi}{4}, -\frac{2\pi}{4}, -\frac{3\pi}{4}$$ Therefore the result is (multiply magnitudes, add phase angles): $$(\sqrt{2})^5 e^{\frac{\pi}{4}i}$$

Answer 3

It think it is a clever approach to calculate both variants. This is a proper way to also verify the calculation. When multiplying the factors it's helpful to group them accordingly.

We obtain \begin{align*} (1+i)&\cdot(i\sqrt{2})(-i\sqrt{2})\cdot(-1+i)(-1-i)\\ &=(1+i)\cdot 2\cdot 2\\ &=4(1+i)\\ &=4\sqrt{2}e^{\frac{\pi}{4}i} \end{align*}

Hint: In the middle group we respect $i\cdot(-i)=1$ and in the right-most group we could think of $(a+b)(a-b)=a^2-b^2$ giving $(-1)^2-(-i)^2=1+1=2$