complex numbers

yen yen · #1 $Old$ November 6th, 2009, 11:39 PM

i got a few questions...
how is (e^3x)sin2x=(e^3x)Im(e^2ix)?
im not very good in complex numbers...

and then
to find d56/dt56 [(e^-t)(sint)]
im confused by
(-1+i)^56=[ √2 e^(3 πi/4)]^56
=(2^28)cos42 π + isin42 π
→(-1+i)^56=2^28 <<<<how do you get this??

Prove It · #2 $Old$ November 6th, 2009, 11:54 PM

Quote:

Originally Posted by yen yen $View Post$

i got a few questions...
how is (e^3x)sin2x=(e^3x)Im(e^2ix)?
im not very good in complex numbers...

and then
to find d56/dt56 [(e^-t)(sint)]
im confused by
(-1+i)^56=[ √2 e^(3 πi/4)]^56
=(2^28)cos42 π + isin42 π
→(-1+i)^56=2^28 <<<<how do you get this??

The first is a direct application of Euler's formula.

A complex number can be written in Cartesian co-ordinates:

$z = \mathbf{Re}(z) + i\mathbf{Im}(z)$

or Polar co-ordinates:

$z = \cos{\theta} + i\sin{\theta} = e^{i\theta}$ .

Can you see that $\cos{\theta} = \mathbf{Re}(z)$ and $\sin{\theta} = \mathbf{Im}(z)$ ?

Can you see that if $\theta = 2x$ then

$\cos{(2x)} + i\sin{(2x)} = e^{2ix}$ ?

So $\sin{(2x)}$ is the imaginary part of $e^{2ix}$ .

yen yen · #3 $Old$ November 6th, 2009, 11:57 PM

oh okay... thanks.. now i get it... still need someone to help me with the second problem...

Prove It · #4 $Old$ November 7th, 2009, 12:21 AM

Quote:

Originally Posted by yen yen $View Post$

i got a few questions...
how is (e^3x)sin2x=(e^3x)Im(e^2ix)?
im not very good in complex numbers...

and then
to find d56/dt56 [(e^-t)(sint)]
im confused by
(-1+i)^56=[ √2 e^(3 πi/4)]^56
=(2^28)cos42 π + isin42 π
→(-1+i)^56=2^28 <<<<how do you get this??

For part 2, are you asking how you go from

$(-1 + i)^{56}$ to $2^{28}$ ?

If so, it's done using DeMoivre's Theorem.

DeMoivre's Theorem states that:

If $z = r\,\textrm{cis}\,{\theta}$

Then $z^n = r^n\,\textrm{cis}\,{(n\theta)}$ for $n \in \mathbf{Z}$ .

So to evaluate $(-1 + i)^{56}$ , convert to polars and use DeMoivre's Theorem.

$r = \sqrt{(-1)^2 + 1^2}$

$= \sqrt{2}$ .

We can see that $\cos{\theta} = -1$ and $\sin{\theta} = 1$ .

The function is in the second quadrant.

So $\frac{\sin{\theta}}{\cos{\theta}} = \frac{1}{-1}$

$\tan{\theta} = -1$

$\theta = \pi - \frac{\pi}{4}$

$= \frac{3\pi}{4}$ .

Therefore

$-1 + i = \sqrt{2}\,\textrm{cis}\,\left(\frac{3\pi}{4}\right)$

Now use DeMoivre's Theorem:

$(-1 + i)^{56} = (\sqrt{2})^{56}\,\textrm{cis}\,\left(\frac{56 \cdot 3\pi}{4}\right)$

$= 2^{28}\,\textrm{cis}\,(42\pi)$ .

Now notice that the angle $42\pi$ is the same as the angle $0$ , since you have just gone around the unit circle 21 times.

So $(-1 + i)^{56} = 2^{28}\,\textrm{cis}\,0$

$= 2^{28}(\cos{0} + i\sin{0})$

$= 2^{28}(1 + 0i)$

$= 2^{28}$ .