Expressing complex function in standard rectangular form

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 7K views
elimenohpee
Messages
64
Reaction score
0
I'm given a complex function in the exponential form:

2.5j e^(-j40*pi)

Transforming this into the standard Cartesian form is pretty straight forward, but the extra j multiplying the 2.5 is kind of throwing me off. I don't know if I did it right, but this is what I did:

2.5j (cos[-40pi] + jsin[-40pi]) = 2.5j (cos[40pi] - jsin[40pi]) = 2.5j (1-0) = 2.5j

Is that correct? I don't know if it is or not, because I don't see how you could transform it back into the original form after putting it into Cartesian form. Thanks
 
Physics news on Phys.org
elimenohpee said:
I'm given a complex function in the exponential form:

2.5j e^(-j40*pi)

Transforming this into the standard Cartesian form is pretty straight forward, but the extra j multiplying the 2.5 is kind of throwing me off. I don't know if I did it right, but this is what I did:

2.5j (cos[-40pi] + jsin[-40pi]) = 2.5j (cos[40pi] - jsin[40pi]) = 2.5j (1-0) = 2.5j

Is that correct? I don't know if it is or not, because I don't see how you could transform it back into the original form after putting it into Cartesian form. Thanks

Looks right to me. Is there a way to check the answer?
 
elimenohpee said:
I'm given a complex function in the exponential form:

2.5j e^(-j40*pi)

Transforming this into the standard Cartesian form is pretty straight forward, but the extra j multiplying the 2.5 is kind of throwing me off. I don't know if I did it right, but this is what I did:

2.5j (cos[-40pi] + jsin[-40pi]) = 2.5j (cos[40pi] - jsin[40pi]) = 2.5j (1-0) = 2.5j

Is that correct? I don't know if it is or not, because I don't see how you could transform it back into the original form after putting it into Cartesian form. Thanks

Yes that's correct :approve:

If you are having difficulty transforming it into the original form, it is probably because the original form is not really the polar form.

The polar form of 2.5j is 2.5e^(j*pi/2) but that is equivalent to the original expression since e^(-j*40pi)=e^0=1 (since the complex exponential has a period of 2pi).
 
Ok good :)

Such an odd question, I guess it was more or less to try and throw you off.