Polar to cartesian form

  • Thread starter SteliosVas
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  • #1
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Homework Statement



Convert 2cis(-pi/3)cis(pi/6) into cartesian form. Show all working to obtain full marks

Homework Equations



I know that the equation for it is 2((cos(theta) +isin(theta))+(cos(theta)+isin(theta)))

The Attempt at a Solution



Okay so cos of (-p/3) = 1/2
Sin of (-p/3) = sqrt(3)/2
Cos(pi/6)=sqrt(3)/2
Sin(pi/6) = 1/2

So basically you get this big answer as 2((1/2-isqrt(3)/2)+(sort(3)/2+1/2i))

Is this the correct answer or do I need to multiply it out?

I know the R for this is 2. I know the theta is tan^-1(y/x) but I am a little confused...
 

Answers and Replies

  • #2
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Homework Statement



Convert 2cis(-pi/3)cis(pi/6) into cartesian form. Show all working to obtain full marks

Homework Equations



I know that the equation for it is 2((cos(theta) +isin(theta))+(cos(theta)+isin(theta)))
In the problem statement you're multiplying, but in the above you're adding. Which is it?
SteliosVas said:

The Attempt at a Solution



Okay so cos of (-p/3) = 1/2
Sin of (-p/3) = sqrt(3)/2
The line above is wrong.
SteliosVas said:
Cos(pi/6)=sqrt(3)/2
Sin(pi/6) = 1/2

So basically you get this big answer as 2((1/2-isqrt(3)/2)+(sort(3)/2+1/2i))

Is this the correct answer or do I need to multiply it out?

I know the R for this is 2. I know the theta is tan^-1(y/x) but I am a little confused...
 
Last edited:
  • #3
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In the problem statement you're multiplying, but in the above you're adding. Which is it?
The above is wrong.
The reason I am adding is because is that not the formula to convert from that form to Cartesian?

In the problem statement you're multiplying, but in the above you're adding. Which is it?
The above is wrong.

What is wrong?
 
  • #4
Ray Vickson
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Homework Statement



Convert 2cis(-pi/3)cis(pi/6) into cartesian form. Show all working to obtain full marks

Homework Equations



I know that the equation for it is 2((cos(theta) +isin(theta))+(cos(theta)+isin(theta)))

The Attempt at a Solution



Okay so cos of (-p/3) = 1/2
Sin of (-p/3) = sqrt(3)/2
Cos(pi/6)=sqrt(3)/2
Sin(pi/6) = 1/2

So basically you get this big answer as 2((1/2-isqrt(3)/2)+(sort(3)/2+1/2i))

Is this the correct answer or do I need to multiply it out?

I know the R for this is 2. I know the theta is tan^-1(y/x) but I am a little confused...
##\text{cis}(a) \times \text{cis}(b) = \text{cis}(a+b)##; see, eg.,http://math.wikia.com/wiki/Cis_θ
 
  • #5
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In the problem statement you're multiplying, but in the above you're adding. Which is it?
The above is wrong.
SteliosVas said:
The reason I am adding is because is that not the formula to convert from that form to Cartesian?
This is what you wrote: "I know that the equation for it is 2((cos(theta) +isin(theta))+(cos(theta)+isin(theta)))"
That is incorrect, and isn't what the formula says.
In the problem statement you're multiplying, but in the above you're adding. Which is it?
The above is wrong.
SteliosVas said:
What is wrong?
You are misunderstanding what the formula says.
 
  • #6
BvU
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The instruction "show all working" leads me to believe they want you to write out the whole thing, thus 'discovering' the wikia product theorem.
Don't work with numbers, just leave the ##-{\pi\over 3}## and the ##\pi \over 6## until the very last.
 
  • #7
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Thanks for that guys, I think I got it now!!!
 

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