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Hello i have some algebra problems. I have the following equation:

$\frac{1}{2}+i\frac{\sqrt{3}}{2}=e^{\Large{\frac{\pi i}{3}}}$

but why is this? and why is the absolute value of this equation equal to $1$?

  • 1
    By Euler's Theorem,$e^{ix}=\cos x+i\sin x$ for all real numbers $x$ – 2017-01-25
  • 1
    The modulus (sometimes called as absolute) of a complex number $a+bi$ is $\sqrt{a^2+b^2}$ – 2017-01-25
  • 0
    @ΘΣΦGenSan Euler's formula also holds for non-real $x$, which gives the complex form of our trig functions. – 2017-01-25
  • 0
    @ Simply Beautiful Art Thanks for the reminder. – 2017-01-25

4 Answers 4

3

Using Euler's theorem, we have that $ e^{x+iy} = e^x(cos(y)+isin(y))$

So in this case, $ e^{\frac{\pi i}{3}} = cos(\frac{\pi}{3}) + isin(\frac{\pi}{3})$.

Since $cos(\frac{\pi}{3}) = \frac{1}{2}$ and $sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$, we see that $ e^{\frac{\pi i}{3}}$ then equals $ \frac{1}{2} + i\frac{\sqrt{3}}{2}$.

The absolute value of a complex number $ x+iy$ is $\sqrt{x^2 + y^2}$, so here the absolute value is $\sqrt{\frac{1}{4} + \frac{3}{4}} = 1 $

1

If we define $e^x$ as $\lim_{n\to\infty}\left(1+\frac xn\right)^n$, then

$$e^{ix}=\lim_{n\to\infty}\left(1+\frac{ix}n\right)^n$$

And while the following image is not rigorous, it does give a good conceptual understanding of why $|e^{ix}|=1$. It also follows that $e^{ix}$ has an angle of $x$ (in radians), and with a bit of trig you could probably conclude that

$$e^{\large\frac{\pi i}3}=\frac12+i\frac{\sqrt3}2$$

(Which should be your $30,60,90$ triangle)

enter image description here

0

Well, we can use:

$$\Re\left(\text{z}\right)+\Im\left(\text{z}\right)i=\sqrt{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}\times e^{\arg\left(\text{z}\right)i}\tag1$$

So:

  • $$\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=\sqrt{\frac{1}{4}+\frac{3}{4}}=\sqrt{\frac{1+3}{4}}=\sqrt{\frac{4}{4}}=1\tag2$$
  • $$\arg\left(\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)=\arctan\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right)=\arctan\left(\sqrt{3}\right)=\frac{\pi}{3}\tag3$$
0

As stated in my comments, the following is known as the Euler's Theorem $$e^{ix}=\cos x+i\sin x $$ for all real numbers $x$.

In your question, $x=\frac{\pi}{3}$. Plugging in, we get

$$e^{i\cdot\frac{\pi}{3}}=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}=\frac{1}{2}+i\frac{\sqrt{3}}{2}.$$ Lastly, the absolute value of $\frac{1}{2}+i\frac{\sqrt{3}}{2}$ is given by $$\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=\sqrt{\frac{1}{4}+\frac{3}{4}}=1.$$