# Complex Numbers; urgent help needed

• Nov 30th 2008, 11:34 AM
14041471
Complex Numbers; urgent help needed
I have spent a considerable amount of time trying to awnser this question. I have no time left... ANY help on this would be much appreciated

We will look at the following complex number. To make it easier, we will call this complex number Z.
http://aventalearning.com/content168...mages/act1.gif

Now find each of these complex numbers. Write the complex number in its usual form (like above I think).

z^2, z^3, z^4...z^9 (I dont expect anyone to solve all, just mabe afew so i can figure out the pattern asuming there is one lol.)

When solving, remember that http://aventalearning.com/content168...mages/act2.gif times http://aventalearning.com/content168...mages/act2.gif equals 1/2 .

• Nov 30th 2008, 11:57 AM
Chris L T521
Quote:

Originally Posted by 14041471
I have spent a considerable amount of time trying to awnser this question. I have no time left... ANY help on this would be much appreciated

We will look at the following complex number. To make it easier, we will call this complex number Z.
http://aventalearning.com/content168...mages/act1.gif

Now find each of these complex numbers. Write the complex number in its usual form (like above I think).

z^2, z^3, z^4...z^9 (I dont expect anyone to solve all, just mabe afew so i can figure out the pattern asuming there is one lol.)

When solving, remember that http://aventalearning.com/content168...mages/act2.gif times http://aventalearning.com/content168...mages/act2.gif equals 1/2 .

Another approach is this:

Note that $z=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i=\frac{1}{ \sqrt{2}}+\frac{1}{\sqrt{2}}i=\cos\left(\tfrac{\pi }{4}\right)+i\sin\left(\tfrac{\pi}{4}\right)=e^{i\ frac{\pi}{4}}$ (by Euler's Formula)

Now:

$z^2=\left[e^{i\frac{\pi}{4}}\right]^2=e^{i\frac{\pi}{2}}=\cos\left(\tfrac{\pi}{2}\rig ht)+i\sin\left(\tfrac{\pi}{2}\right)=i$

$z^3=\left[e^{i\frac{\pi}{4}}\right]^3=e^{i\frac{3\pi}{4}}=\cos\left(\tfrac{3\pi}{4}\r ight)+i\sin\left(\tfrac{3\pi}{4}\right)=-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$

....

$z^9=\left[e^{i\frac{\pi}{4}}\right]^9=e^{i\frac{9\pi}{4}}=\cos\left(\tfrac{9\pi}{4}\r ight)+i\sin\left(\tfrac{9\pi}{4}\right)=\frac{1}{\ sqrt{2}}+\frac{1}{\sqrt{2}}i$, which happens to be $z$

Does this make sense?
• Nov 30th 2008, 12:05 PM
Soroban
Hello, 14041471!

Are you familiar with DeMoivre's Theorem: . $(\cos\theta + i\sin\theta)^n \;=\;\cos(n\theta) + i\sin(n\theta)$

Quote:

$z \:=\:\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$

Find: . $z^2,\;z^3,\;z^4,\;\hdots z^9$

We have: . $z \;=\;\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4}$

$z^2 \;=\;\left(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4}\right)^2 \;=\;\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2} \;=\;0 + i \;=\;i$

$z^3 \;=\;\cos\tfrac{3\pi}{4} + i\sin\tfrac{3\pi}{4} \;=\;\text{-}\tfrac{\sqrt{2}}{2} + \tfrac{\sqrt{2}}{2}i$

$z^4 \;=\;(z^2)^2 \;=\;(i)^2 \;=\;-1$

$z^5 \;=\;(z^4)(z) \;=\;(-1)\left(\tfrac{\sqrt{2}}{2} + \tfrac{\sqrt{2}}{2}i\right) \;=\;\text{-}\tfrac{\sqrt{2}}{2}- \tfrac{\sqrt{2}}{2}i$

$z^6 \;=\;(z^2)^3 \;=\;i^3 \;=\;-i$

Get the idea?