Modulus of complex numbers z=(cos+i sin)^7
- Thread starter Lula
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Hey
So I have a complex number written in the form z=(cos + i sin)^7 which is not how I’m used to seeing them written. I would usually get a problem that looks like z^7=(cos + i sin)^7 where both sides are raised to the power, so I wondered how the method of finding the modulus would change for numbers in this form.
to find the modulus I would usually convert to rectangular form [z=(x+yi)] and then (x^2+y^2)^1/2 = mod Z
So I have a complex number written in the form z=(cos + i sin)^7 which is not how I’m used to seeing them written. I would usually get a problem that looks like z^7=(cos + i sin)^7 where both sides are raised to the power, so I wondered how the method of finding the modulus would change for numbers in this form.
to find the modulus I would usually convert to rectangular form [z=(x+yi)] and then (x^2+y^2)^1/2 = mod Z
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"pretty sure" - Not a good foundation for mathematics.
You should have met Euler's Formula: [math]e^{it} = r\cdot\left(cos(t)+i\cdot\sin(t)\right)[/math]
Use your oldest algebra to imagine what exponentiation will do to the modulus of both sides of that.
You should have met Euler's Formula: [math]e^{it} = r\cdot\left(cos(t)+i\cdot\sin(t)\right)[/math]
Use your oldest algebra to imagine what exponentiation will do to the modulus of both sides of that.
D
Deleted member 4993
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You write:
z=(cos + i sin)^7 .........................that is incorrect. It probably should have been:
z=(cos(t) + i sin(t))^7
Please do not double post.
topsquark
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Pet peeve warning!!
You can't just write sin and cos. These are functions which have an argument, like [math]sin( \theta )[/math] and [math]cos( \theta )[/math]. These are needed to give the trig. functions meaning.
If you can't/won't use something like [math]\theta[/math], feel free to simply use a t if you can. But you have to have something there.
End of pet peeve!
-Dan
You can't just write sin and cos. These are functions which have an argument, like [math]sin( \theta )[/math] and [math]cos( \theta )[/math]. These are needed to give the trig. functions meaning.
If you can't/won't use something like [math]\theta[/math], feel free to simply use a t if you can. But you have to have something there.
End of pet peeve!
-Dan
pka
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If \(z=a+bi\) then the modulus is \(|z|=\sqrt{a^2+b^2}\). For any \(n\) it is the case that \(|z^n|=|z|^n\).
Please note that for any \(\theta\), the modulus \(|\cos(\theta)+i\sin(\theta)|=1\)
To find polar form you need to find the argument:
\(\arg(x + yi) = \left\{ {\begin{array}{{rl}} {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right.\)
Ok so I can find the modulus in whatever for
okay that’s a good point but do you have any suggestions for the question? Do you think I should go about solving this starting with rearranging to
Z^(1/7) = cos(t)+i sin(t) ?
okay that’s a good point but do you have any suggestions for the question? Do you think I should go about solving this starting with rearranging to
Z^(1/7) = cos(t)+i sin(t) ?
I have a number in the form
Z= [cos(t) - i sin(t)]^7
and I’m trying to find the modulus and argument.
Would it be correct to re-write as
Z^(1/7) =[cos(t) - i sin(t)]
and go from there or would it be correct to re-write as
z=7[cos(7t) - i sin(7t)]
Z= [cos(t) - i sin(t)]^7
and I’m trying to find the modulus and argument.
Would it be correct to re-write as
Z^(1/7) =[cos(t) - i sin(t)]
and go from there or would it be correct to re-write as
z=7[cos(7t) - i sin(7t)]
Last edited:
D
Deleted member 4993
Guest
Have you studied:
De Moivre's formula
You need to apply that here. If you have not heard of it before - do a google search, there are excellent video and "written" lessons are available.
Study those. If you are still stuck - come back and tell us exactly where you got stuck. We can unstuck you...
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Quite the contrary. There are many places to to seek and receive instruction, level of confidence notwithstanding.
pka
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I have shown you in reply #8 that \(|\cos(t)-i\sin(t)|=1\) thus \(|\cos(t)-i\sin(t)|^7=1\)
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