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.
$\endgroup$ 02 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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
$|y|=1$
$y^6=\frac{1}{2}-\frac{\sqrt{3}}{2}i$
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
$-\frac{1}{c}2^{50}$
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