# Complex Number Help

• Aug 29th 2012, 12:15 AM
timeforchg
Complex Number Help

Solve the equation : http://www.askmehelpdesk.com/cgi-bin...281+2j%29]%5E6 giving your answers in the form of a + jb. Show that there can be only be 8 distinct roots.
• Aug 29th 2012, 05:23 AM
Soroban
Re: Complex Number Help
Hello, timeforchg!

Where did this problem come from?
Do you have the answers? .Any hints?

Is there a typo?
If the right side is a constant, there are only six roots.

Quote:

$\displaystyle \text{Solve the equation: }\:z^6-1 \;=\;\left[\frac{64j(1-j)}{1+2j}\right]^6$

. . $\displaystyle \text{giving your answers in the form }a+ib.$

$\displaystyle \text{Show that there can be only be 8 distinct roots.}$

Consider the fraction: .$\displaystyle \frac{64j(1-j)}{1+2j} \;=\;\frac{64(j - j^2)}{1+2j} \;=\; \frac{64(1 + j)}{1+2j}$

Rationalize the denominator: .$\displaystyle \frac{64(1+j)}{1+2i}\cdot\frac{1-2j}{1-2j} \;=\; \tfrac{64}{5}(3-j)$

The equation becomes: .$\displaystyle z^6-1 \;=\; \left[\tfrac{64}{5}(3-j)\right]^6 \quad\Rightarrow\quad z^6 \;=\;\left[\tfrac{64}{5}(3-j)\right]^6 + 1$

. . . . . . . . . . . . . . . . . . . . . $\displaystyle z \;=\; \sqrt[6]{\left[\tfrac{64}{5}(3-j)\right]^6 + 1}$
Ugly!
• Aug 29th 2012, 06:05 AM
Prove It
Re: Complex Number Help
Quote:

Originally Posted by timeforchg

Solve the equation : http://www.askmehelpdesk.com/cgi-bin...281+2j%29]%5E6 giving your answers in the form of a + jb. Show that there can be only be 8 distinct roots.

It's ugly, but not unmanageable...

\displaystyle \displaystyle \begin{align*} z^6 - 1 &= \left[\frac{64j(1-j)}{1+2j}\right]^6 \\ z^6 - 1 &= \left(\frac{64+64j}{1 + 2j}\right)^6 \\ z^6 - 1 &= \left[\frac{(64 + 64j)(1 - 2j)}{(1 + 2j)(1 - 2j)}\right]^6 \\ z^6 - 1 &= \left( \frac{192}{5} - \frac{64}{5}j \right)^6 \\ z^6 - 1 &= \left[\frac{64\sqrt{10}}{5}\,e^{j\left( \pi - \arctan{\frac{1}{3}}\right)}\right]^6 \\ z^6 - 1 &= \frac{ 2^{36}\cdot 10^3 }{5^6}\,e^{6j\left( \pi - \arctan{\frac{1}{3}}\right)} \\ z^6 - 1 &= \frac{ 2^{39} }{5^3}\,e^{6\pi j}\,e^{j\left(-6\arctan{\frac{1}{3}}\right)} \\ z^6 - 1 &= \frac{2^{39}}{5^3} \, e^{j\left(-6\arctan{\frac{1}{3}}\right)} \\ z^6 - 1 &= \frac{2^{39}}{5^3}\left(-\frac{44}{125} - \frac{117}{125}j \right) \\ z^6 - \frac{5^6}{5^6} &= -\frac{2^{41} \cdot 11}{5^6} - \frac{2^{39}\cdot 3^2 \cdot 13}{5^6}j \\ z^6 &= \frac{5^6 - 2^{41} \cdot 11}{5^6} - \frac{2^{39} \cdot 3^2 \cdot 13}{5^6}j \end{align*}

Now by converting the RHS to the most general polar form and taking both sides to the power of 1/6 will yield the desired answer.
• Sep 3rd 2012, 05:27 PM
timeforchg
Re: Complex Number Help
Hi Guys,

Sorry there was a typo. Yes it is 6 distinct roots. Thanks a lot for the head start.
• Sep 13th 2012, 07:38 PM
MaxJasper
Re: Complex Number Help
$\displaystyle {z6}=1+\left(\frac{64 i (1-i)}{1+2 i}\right)^6$

$\displaystyle z6 = ((64 i (1 - i))/(1 + 2 i))^6 + 1$

$\displaystyle {zz}({k}){=}\sqrt[6]{|{z6}|} e^{i \arg \left(\frac{2 \pi k}{6}+\frac{\text{z6}}{6}\right)}$

$\displaystyle k = {0, 1, 2, 3, 4, 5}$

z1, z2, z3, z4, z5, z6=
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{7146825580544}{2687695088383}\right) }$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-31250 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-62500 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{21440476741632}{8063085265149-31250 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-125000 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-156250 \pi }\right)}$
• Sep 16th 2012, 07:17 AM
timeforchg
Re: Complex Number Help
Thanks for the help guys.

Hi Proveit,

A quick question. how do you derive from equation A to B. Pls See attached. :)Attachment 24830
• Sep 16th 2012, 07:37 AM
Prove It
Re: Complex Number Help
Expanded the brackets in the power, then applied \displaystyle \displaystyle \begin{align*} a^{m+n} = a^ma^n \end{align*}.
• Sep 16th 2012, 06:39 PM
timeforchg
Re: Complex Number Help
Quote:

Originally Posted by Prove It
Expanded the brackets in the power, then applied \displaystyle \displaystyle \begin{align*} a^{m+n} = a^ma^n \end{align*}.

I see. Correct me if I am wrong. Do we need the same base in order to apply that theorem?
• Sep 16th 2012, 08:45 PM
Prove It
Re: Complex Number Help
Yes, and the base in question is e.
• Sep 18th 2012, 11:25 AM
hiroyoki
Re: Complex Number Help
hi jasper i read the question i dont really get what u mean on the part zz(K)=
• Sep 18th 2012, 05:09 PM
timeforchg
Re: Complex Number Help
Quote:

Originally Posted by hiroyoki
hi jasper i read the question i dont really get what u mean on the part zz(K)=

hmmm... someone interested in this question uh.. from SG some more. haha
• Sep 18th 2012, 05:13 PM
xotinalx
Re: Complex Number Help
hmm
• Sep 18th 2012, 06:33 PM
hiroyoki
Re: Complex Number Help
SO timeforchg u understand that part?
• Sep 23rd 2012, 04:58 PM
timeforchg
Re: Complex Number Help
SOLVED!! Thanks a lot guys!! Appreciate it.