MHB How does theta transform into theta minus pi/2 in complex numbers?

[aruwin]

Hi!
Can you tell me how the theta changes into theta minus pi/2? Can you show me, please?

 

[MarkFL]

Hint: Use Euler's formula and the co-function identities for sine and cosine to demonstrate that:

$$ie^{-i\theta}=e^{i\left(\frac{\pi}{2}-\theta\right)}$$

 

[aruwin]

Hint: Use Euler's formula and the co-function identities for sine and cosine to demonstrate that:

$$ie^{-i\theta}=e^{i\left(\frac{\pi}{2}-\theta\right)}$$

How did the "i" disappear?

 

[Nono713]

How did the "i" disappear?

$i = e^{i \frac{\pi}{2}}$

 

[chisigma]

The problem with this 'identity' is that in the first term there is the factor $\displaystyle e^{- j\ \theta} $ and in second there is the factor $\displaystyle e^{j\ \theta}$... and that is impossible!(Tmi)...

Kind regards

$\chi$ $\sigma$

 

[aruwin]

The problem with this 'identity' is that in the first term there is the factor $\displaystyle e^{- j\ \theta} $ and in second there is the factor $\displaystyle e^{j\ \theta}$... and that is impossible!(Tmi)...

Kind regards

$\chi$ $\sigma$

Do you mean that the equation is actually wrong?

 

[Prove It]

The problem with this 'identity' is that in the first term there is the factor $\displaystyle e^{- j\ \theta} $ and in second there is the factor $\displaystyle e^{j\ \theta}$... and that is impossible!(Tmi)...

Kind regards

$\chi$ $\sigma$

Except that the second term is out by a factor of "i", which DOES make it correct.

 

[chisigma]

Do you mean that the equation is actually wrong?

Yes, I do... the correct procedure in my opinion is...

$\displaystyle \frac{E_{s}^{2} - E_{s}\ E_{r}\ e^{- j\ \theta}}{- j\ X} = j\ \frac{E_{s}^{2}}{X} - j\ \frac{E_{s}\ E_{r}}{X}\ e^{- j\ \theta} = j\ \frac{E_{s}^{2}}{X} + \frac{E_{s}\ E_{r}}{X}\ e^{- j\ (\theta + \frac{\pi}{2})}$

Kind regards

$\chi$ $\sigma$

 

[aruwin]

Yes, I do... the correct procedure in my opinion is...

$\displaystyle \frac{E_{s}^{2} - E_{s}\ E_{r}\ e^{- j\ \theta}}{- j\ X} = j\ \frac{E_{s}^{2}}{X} - j\ \frac{E_{s}\ E_{r}}{X}\ e^{- j\ \theta} = j\ \frac{E_{s}^{2}}{X} + \frac{E_{s}\ E_{r}}{X}\ e^{- j\ (\theta + \frac{\pi}{2})}$

Kind regards

$\chi$ $\sigma$

Could you explain to me how the j on the right disappeared?

 

[chisigma]

Could you explain to me how the j on the right disappeared?

... because is $\displaystyle j= e^{\ j\ \frac{\pi}{2}}$...

Kind regards

$\chi$ $\sigma$

 

[MarkFL]

How did the "i" disappear?

This is what I had in mind:

Use Euler's formula:

$$ie^{-i\theta}=i\left(\cos(\theta)-i\sin(\theta\right)$$

Use co-function identities:

$$ie^{-i\theta}=i\left(\sin\left(\frac{\pi}{2}-\theta\right)-i\cos\left(\frac{\pi}{2}-\theta\right)\right)$$

Distribute the $i$ and rearrange:

$$ie^{-i\theta}=\cos\left(\frac{\pi}{2}-\theta\right)+i\sin\left(\frac{\pi}{2}-\theta\right)$$

Use Euler's formula:

$$ie^{-i\theta}=e^{i\left(\frac{\pi}{2}-\theta\right)}$$