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What is the norm of a complex exponential?
After class today, a student asked me the following question:
$ \left| e^{j \omega} \right| = ? $
Please help answer this question.
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
$ e^{j \omega} = cos( \omega) + i*sin( \omega) $
hence,
$ \left| e^{j \omega} \right| = \left|cos( \omega) + i*sin( \omega) \right| = \sqrt{cos^2( \omega) + sin^2( \omega)} = 1 $
- TA's comments: Is this true for all $ \omega \in R $? The answer is yes.
- Instructor's comment: I would like to propose a more straightforward way to compute this norm using the fact that $ |z|^2=z \bar{z} $. Can you try it out? -pm
Answer 2
becasue: $ e^{jx} =cos(x)+ jsin(x) $
$ | e^{j \omega}|=|cos(\omega) + i*sin(\omega)|=\sqrt{cos(\omega)^2 +sin(\omega)^2}=1 $
- TA's comments: The point here is to use Euler's formula to write a complex exponential as a complex number. Then the norm(magnitude) and angle(phase) of this complex number can be easily computed.
- Instructor's comment: Again, I would argue that using the fact that $ |z|^2=z \bar{z} $ is more straightforward. Can you try it out? -pm
Answer 3
$ e^{j \omega} = cos( \omega) + i*sin( \omega) $
$ \left| e^{j \omega} \right| = \left|cos( \omega) + i*sin( \omega) \right| = \sqrt{cos^2( \omega) + sin^2( \omega)} = 1 $
- Instructor's comment: Can you think of a way to compute this norm without using Euler's formula? -pm
