Principal arguements

Nov 2009
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I've got to find the modulus and principle argument and I'm a little unsure how to do it for this example:

cos(pi/7) + isin(pi/7)

I can do it perfectly on complex numbers, but I'm unsure how to start this one?

Thanks :]
 
Oct 2009
4,261
1,836
I've got to find the modulus and principle argument and I'm a little unsure how to do it for this example:

cos(pi/7) + isin(pi/7)

I can do it perfectly on complex numbers, but I'm unsure how to start this one?

Thanks :]

What do you mean by "I can do it perfectly on complex numbers, but I'm unsure how to start this one"?? You're given a complex number!

\(\displaystyle z=\cos \pi/7+i\sin \pi/7\Longrightarrow Re(z)=\cos \pi/7\,,\,\,Im(z)=\sin \pi/7\) \(\displaystyle \Longrightarrow |z| = \sqrt{Re(z)^2+Im(z)^2}=1\,,\,\,\arg(z)=\arctan Im(z)/Re(z)=\pi/7\) ....

Tonio
 
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HallsofIvy

MHF Helper
Apr 2005
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(Giggle)
I've got to find the modulus and principle argument and I'm a little unsure how to do it for this example:

cos(pi/7) + isin(pi/7)

I can do it perfectly on complex numbers, but I'm unsure how to start this one?

Thanks :]
I should read through the entire thread before I respond! Tonio answer this half an hour ago and in almost the same words! (For some reason, this thread is still listed as having "0 responses".)

I am confused. If you cam "do it perfectly on complex numbers" why can't you do it for this complex number?

Do you mean you can find the modulus and principle argument for something like a+ bi? I presume, then, that you know the modulus is \(\displaystyle \sqrt{a^2+ b^2}\) and the "principle argument" is the principle value for the arctangent function, \(\displaystyle arctan(\frac{b}{a})\).

Then how could you have a problem here? With \(\displaystyle a= cos(\pi/7)\) and \(\displaystyle b= sin(\pi/7)\) you should be able to see immediately that \(\displaystyle \sqrt{a^2+ b^2}= \sqrt{cos^2(\pi/7)+ sin^2(\pi/7)}= 1\) because \(\displaystyle sin^2(\theta)+ cos^2(\theta)= 1\) for all \(\displaystyle \theta\).

And surely you can do \(\displaystyle arctan(\frac{b}{a})= arctan(\frac{sin(\pi/7)}{cos(\pi/7)})= arctan(tan(\pi/7))\).

I think the difficulty is that this problem is too easy!
 
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