MHB What is the Result of Raising a Complex Number to a Power?

  • Thread starter Thread starter Petrus
  • Start date Start date
  • Tags Tags
    Form Rectangular
Click For Summary
The discussion focuses on calculating the expression \(\left(\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)^{100}\) in the form \(a+ib\). The initial calculations correctly identify the modulus and argument, leading to the expression \(1*e^{i\frac{100\pi}{3}}\). However, a mistake is made in simplifying the angle, which should be \(\frac{4\pi}{3}\) instead of \(\frac{\pi}{3}\). Applying de Moivre's theorem reveals that the correct result is \(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\). The discussion highlights the importance of accurately calculating angles in complex number exponentiation.
Petrus
Messages
702
Reaction score
0
Hello MHB,
calculate $$\left(\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)^{100}$$ in the form $$a+ib$$

progress:
I start to calculate argument and get it to $$r=1$$ (argument)
then $$\cos\theta=\frac{1}{2} \ sin\theta=\frac{\sqrt{3}}{2}$$ we se it's in first quadrant( where x and y is positive)
$$1*e^{i\frac{100\pi}{3}}$$
notice that we can always take away 2pi so we can simplify that to
$$1*e^{i\frac{\pi}{3}}$$
$$1*e^{i\frac{\pi}{3}}=\cos(\frac{\pi}{3}) + i \sin (\frac{\pi}{3}) = \frac{1}{2}+i\frac{\sqrt{3}}{2}$$
but the facit says $$\frac{-1}{2}-i\frac{\sqrt{3}}{2}$$

Regards,

 
Mathematics news on Phys.org
Petrus said:
Hello MHB,
calculate $$\left(\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)^{100}$$ in the form $$a+ib$$

progress:
I start to calculate argument and get it to $$r=1$$ (argument)
then $$\cos\theta=\frac{1}{2} \ sin\theta=\frac{\sqrt{3}}{2}$$ we se it's in first quadrant( where x and y is positive)
$$1*e^{i\frac{100\pi}{3}}$$
notice that we can always take away 2pi so we can simplify that to
$$1*e^{i\frac{\pi}{3}}$$
$$1*e^{i\frac{\pi}{3}}=\cos(\frac{\pi}{3}) + i \sin (\frac{\pi}{3}) = \frac{1}{2}+i\frac{\sqrt{3}}{2}$$
but the facit says $$\frac{-1}{2}-i\frac{\sqrt{3}}{2}$$

Regards,

Hey Petrus!

What is $$\frac {100\pi}{3} \pmod{2\pi}$$?
 
I like Serena said:
Hey Petrus!

What is $$\frac {100\pi}{3} \pmod{2\pi}$$?
$$\frac{4}{3}$$

- - - Updated - - -

Thanks I like Serena I see what I did wrong :)

Regards,
 
Petrus said:
$$\frac{4}{3}$$

That should be $$\frac{4\pi}{3}$$. (Yeah, I know, I'm a nitpicker.)

So what's $$\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}$$?
 
I like Serena said:
(Yeah, I know, I'm a nitpicker
nitpicker or not, I am grateful for the fast responed!

Regards,
 
If I were to work the problem, I would write:

$$\left(\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)^{100}= \left(\cos\left(\frac{\pi}{3}+2k\pi \right)+i\sin\left(\frac{\pi}{3}+2k\pi \right) \right)^{100}$$

Now applying de Moivre's theorem we have:

$$\cos\left(\frac{100\pi}{3}+200k\pi \right)+i\sin\left(\frac{100\pi}{3}+200k\pi \right)=\cos\left(\frac{4\pi}{3}+232k\pi \right)+i\sin\left(\frac{4\pi}{3}+232k\pi \right)= -\left(\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K