Expressing complex function in standard rectangular form

[elimenohpee]

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

 

[berkeman]

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?

 

[gabbagabbahey]

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

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).

 

[elimenohpee]

Ok good :)

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