Express the following complex numbers in the form of $a+bi$

Question

I'm struggling with the following exercises:

I tried to use the reasoning as follows:

$$(a+bi)^n=(re^{\theta i})^n=r^ne^{\theta in}=r^n(\cos(\theta n)+i\sin(\theta n))$$

So for the first one I did:

$$2^{1/6}(\cos(-\frac{\pi}{3} \cdot \frac{1}{6})+i\sin(-\frac{\pi}{3} \cdot \frac{1}{6}))$$

But it gives me a decimal solution, that's why I'm not sure about the solution.

And for the second one, I was attempting to do the same but when I was calculating $r$ of $(1+\sqrt{-3}i)^{50}$, that is, the modulus of the complex number:

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

which doesn't exist.

Any idea? Thank you.

Answer 1

The first one is correct. $a$ and $b$ can be decimal numbers, no worries...

For the second one, assuming that it's not a mistake of the author of the question, then $\sqrt{-3}$ may be interpreted as $i\sqrt{3}$ since $(i\sqrt{3})^2 = -3$. Thus $1+\sqrt{-3}i = 1 + i\sqrt{3}.i = 1-\sqrt{3}$ which is real.

Answer 2

We can calculate the norm of $z$ in this way:

$|z|^6=|z^6|=|1-\sqrt{3}i|=2$ so you have that

$|z|= 2^\frac{1}{6}$

Now you can define $y:= \frac{z}{2^\frac{1}{6}}$ and then

1. $|y|=1$

2. $y^6=\frac{1}{2}-\frac{\sqrt{3}}{2}i$

You can write $y$ in trigonometric form:

$y=cos(\theta)+i sin(\theta)$

and so

$\frac{1}{2}-\frac{\sqrt{3}}{2}i =y^6=cos(6\theta)+i sin(6\theta)$

Then you must impose the conditions:

$cos(6\theta)=\frac{1}{2}$

$sin(6\theta)=-\frac{\sqrt{3}}{2}$

and the solution is

$\theta=\frac{-\pi}{36}+\frac{k\pi}{3}$ for $k\in\{-1,-2,0,1,2,3\}$

for example for $k=0$ you have that

$y=cos(\frac{\pi}{36})-i sin(\frac{\pi}{36})$

and so

$z=2^{\frac{1}{6}}(cos(\frac{\pi}{36})-i sin(\frac{\pi}{36}))$

For the second

$(1+\sqrt{-3}i)^{50}=(1-\sqrt{3})^{50}=c$ that is a real number

$(1+i)^{100}=2^{50}(cos(\frac{100\pi}{4})+ i sin(\frac{100\pi}{4}))=$

$ 2^{50}(-1+ i 0)=-2^{50}$

So the number is real and is

$-\frac{1}{c}2^{50}$