MHB Complex numbers finding the Imaginary part

[Drain Brain]

is there a way to solve this without performing the tedious expansion of $(1+j)^8$?

here's the problem

$\text{Im}[(1+j)^8(x+jy)]$

 

[Prove It]

is there anyway to solve this without performing the tedious expansion of $(1+j)^8$?

here's the problem

$\text{Im}[(1+j)^8(x+jy)]$

Exponentiation of complex numbers is easiest if it's in its exponential polar form...

 

[Drain Brain]

Exponentiation of complex numbers is easiest if it's in its exponential polar form...

how to do that polar form?

 

[I like Serena]

how to do that polar form?

The polar form is:
$$1+j=\sqrt 2 \cdot e^{j \pi / 4}$$

Generally it is:
$$x+yj = r e^{j\phi}$$
where $x= r\cos \phi$ and $y=r\sin \phi$.

 

[Drain Brain]

The polar form is:
$$1+j=\sqrt 2 \cdot e^{j \pi / 4}$$

Generally it is:
$$x+yj = r e^{j\phi}$$
where $x= r\cos \phi$ and $y=r\sin \phi$.

Is there another way aside from that? It seems I can't handle that yet.

 

[I like Serena]

Is there another way aside from that? It seems I can't handle that yet.

Well... we can apply the binomium:
$$(a+b)^n = a^n + \binom n 1 a^{n-1}b + \binom n 2 a^{n-2}b^2 + ... + b^n$$
In your case that would be:
$$(1+j)^8 = 1 + 8j + \binom 8 2 j^2 + ... + j^8 = 1 + 8j - \binom 8 2 - \binom 8 3 j +\binom 8 4 + ... + 1$$And let me show how it's done using polar:
$$(1+j)^8 = \Big(\sqrt 2 e^{j \pi /4}\Big)^8 = (\sqrt 2)^8 e^{2\pi j} = (\sqrt 2)^8 \cdot 1 = (\sqrt 2)^8$$Otherwise I'm afraid you're stuck with expanding the expression factor by factor.

 

[Nono713]

If you don't want to use polar form, it's not that tedious to expand the product using the binomial theorem:
$$(1 + j)^8 = \sum_{k = 0}^8 \binom{8}{k} j^k$$
Now recall the cyclic property of exponents of $j$ that you saw in the other thread, then you will know those nine terms, in terms of $k$, will fall into exactly two equivalence classes in terms of real and imaginary parts, given by $(0, 2, 4, 6, 8)$ and $(1, 3, 5, 7)$, in other words, even $k$ produces a real term and odd $k$ produces an imaginary term with alternating signs ($0 \to 1$, $1 \to i$, $2 \to -1$, $3 \to -i$ and so on). We calculate each separately:
$$\Re((1 + j)^8) = \binom{8}{0} - \binom{8}{2} + \binom{8}{4} - \binom{8}{6} + \binom{8}{8} = 16$$
$$\Im((1 + j)^8) = \binom{8}{1} - \binom{8}{3} + \binom{8}{5} - \binom{8}{7} = 0$$
The imaginary part is easy since the binomial terms cancel out due to the property $\binom{n}{k} = \binom{n}{n - k}$. For the real part I guess you just have to compute each term, it's not that bad, there's only two nontrivial terms to work out since $\binom{n}{0} = 1$. So we conclude $(1 + j)^8 = 16 + 0j = 16$. The rest follows...​

 

[Deveno]

Often, the complex numbers are called "rotation-dilations" because multiplying the vector $(x,y) = x+yj$ by $a+bj$ corresponds to the linear map $\Bbb R^2 \to \Bbb R^2$ with matrix:

$\begin{bmatrix}a&-b\\b&a \end{bmatrix}$.

This stretches the vector $(x,y)$ by a factor of $\sqrt{a^2 + b^2}$ and rotates it through the angle:

$\theta = \tan^{-1}\left(\dfrac{b}{a}\right)$ if $a > 0$
$\theta = \dfrac{\pi}{2}$, if $a = 0, b > 0$
$\theta = \tan^{-1}\left(\dfrac{b}{a}\right) + \pi$ if $a < 0, b \geq 0$
$\theta = \tan^{-1}\left(\dfrac{b}{a}\right) - \pi$ if $a < 0, b < 0$
$\theta = -\dfrac{\pi}{2}$, if $a = 0, b < 0$

(the rotation induced by the 0-vector is left undefined).

The important thing to remember about this is: when we multiply complex numbers, the magnitudes multiply, and the angles add.

So what does this mean for $(1+j)^8$?

Note that this lies on the line $y = x$ with angle $\tan^{-1}(1) = \dfrac{\pi}{4}$ (45 degrees). So when we take it to the 8th power, we are going to get a vector with angle $2\pi$, which is the same as a vector with angle 0: that is, it lies on the $x$ (real) axis, pointing in the positive direction.

We are also going to take the magnitude of $1 + j$ to the 8th power: so we need to find this. But it's clear that $1+j$ is the hypotenuse of an isosceles right triangle with equal legs of length 1, and so has magnitude $\sqrt{2}$. Thus our magnitude will be:

$(\sqrt{2})^8 = ((\sqrt{2})^2)^4 = 2^4 = 16$

so $(1+j)^8$ is a vector with magnitude 16 pointing in the positive direction of the $x$-axis, that is to say: the real number 16.