Hi guys, I am totally lost in my class and am not sure about this tutorial. Help please...
Solve the equation : giving your answers in the form of a + jb. Show that there can be only be 8 distinct roots.
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.
$\displaystyle \text{Solve the equation: }\:z^6-1 \;=\;\left[\frac{64j(1-j)}{1+2j}\right]^6$
. . $\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!
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.
$\displaystyle {z6}=1+\left(\frac{64 i (1-i)}{1+2 i}\right)^6$
$\displaystyle z6 = ((64 i (1 - i))/(1 + 2 i))^6 + 1$
$\displaystyle {zz}({k}){=}\sqrt[6]{|{z6}|} e^{i \arg \left(\frac{2 \pi k}{6}+\frac{\text{z6}}{6}\right)}$
$\displaystyle k = {0, 1, 2, 3, 4, 5}$
z1, z2, z3, z4, z5, z6=
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{7146825580544}{2687695088383}\right) }$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-31250 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-62500 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{21440476741632}{8063085265149-31250 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-125000 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-156250 \pi }\right)}$