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I am doing EE hw right now while going over my notes, I notice that my prof said that e^(jpi/4) = 1 but how? Using Euler's, I would get e(jpi/4) = cos(pi/4) + jsin(pi/4) = 0.7071 + j0.7071 enter image description here

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  • \$\begingroup\$ Take the magnitude of cos(pi/4) + jsin(pi/4) and you'll get one. x+iy is a vector, it has direction and magnitude. 1 is just the magnitude only, 0.7+0.7j has a magnitude of one and an angle of 45'. Magnitude is sqrt(x^2 + y^2). \$\endgroup\$ – Sam May 5 '16 at 9:45
  • \$\begingroup\$ Those vertical lInes mean magnitude \$\endgroup\$ – Scott Seidman May 5 '16 at 12:08
  • \$\begingroup\$ The edits you have made to your question are wrong for two reasons. 1) you have tried to correct your numbers after I pointed out the problem and 2) the picture you have added (since my answer) is irrelevant. If you are going to edit a question following what is revealed in an answer, make sure you do it in a way that people reading my answer will not be confused into thinking I've missed the point. Common courtesy really. \$\endgroup\$ – Andy aka May 5 '16 at 13:57
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enter image description here

I don't know how you managed to get cos(pi/4) = 0.5 because cos(45) = 0.7071 and sin(45) = 0.7071 too. (These errors were subsequently edited out of the question).

Hence \$\sqrt{0.7071^2 + 0.7071^2}\$ = 1

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I would like to point out that complex exponentials are really another way to write complex numbers. So if there is no coefficient of exp(j*omega) then the magnitude is implied to be 1. The coefficient IS the magnitude.

$$ z = \alpha + j\beta = r e^{j\omega} $$ $$ |z| = \sqrt{\alpha^2+\beta^2} =r$$ $$ \angle z = \arg(z)=\omega=\arctan(\frac{\beta}{\alpha}) $$

Hope this helps

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