# Simple question about complex $e^{i}$ and angles

I'm working with angles.

I have a hard time figuring something. In electric physics, I have an equation describing an AC voltage function, this way

$V_{x} = 0.0469 \cdot e^{-j \cdot 1.083}\cdot e^{j(200\pi \cdot t)}$

Well, I can't get why it does in the solutions. Indeed when they derive it it does :

$\frac{ d\{V_{x}\}}{dt} = 29.47 \cdot e^{0.488\cdot j}\cdot e^{j(200\pi \cdot t)}$

j is equal to $\sqrt{-1}$, it's the standard 'i'.

How can we pass from $e^{-1.083j}$ directly to positive $e^{0.488j}$ is a mystery for me. I've tried to resolve it again and again. Well... I'm stuck. -1.083 and 0.488 are angular displacement. I know we must try to obtain always the lowest displacement. Still... I'm confused about this one.

Can you help?

Thanks

## 1 Answer

Firstly, did you get the following for your derivative? $$\frac{dV_x}{dt} = 0.0469e^{-1.083j}e^{(200\pi t)j}(200\pi j).$$ If I write $j$ as: $$j=e^{j\pi/2}.$$ Then we can take the following two terms from the derivative: $$je^{-1.083j}=e^{j\pi/2}e^{-1.083j}=e^{j(\pi/2 - 1.083)}$$

Now use the following approximation: $\pi/2\approx 1.5708$, which gives $$j(\pi/2 - 1.083)=j(1.5708-1.083)=0.4878j \approx 0.488j$$

I think I have shown all the necessary steps.

• Ohhhhhhhhhhh........... I had forgotten to put $200\pi j$ when I derived. I did multiply with $200 \pi$. The most funny is that I've been starring at my desk on this thing for an hour. Thanks a lot. – Yannick Nov 8 '13 at 7:52
• I thought it was most likely a problem with the $j$ introduced in the derivative! – Zephos Nov 8 '13 at 8:05