Pierwiastki zespolone rownania

[profesorq]

Znajdz wsyzstkie zespolone pierwiastki równania
\(\displaystyle{ (z^2+13)(z^3-\frac{27-54i}{2+i})=0}\)

[soku11]

Pierwszy element:
\(\displaystyle{ z^2=-13=13(cos\pi+isin\pi)\\
z_k=\sqrt{13}(cos\frac{\pi+2k\pi}{2}+isin\frac{\pi+2k\pi}{2})\ \ k\in\{0,1\}\\}\)

Drugi element:
\(\displaystyle{ \frac{27-54i}{2+i}=
\frac{(27-54i)(2-i)}{4-i^2}=
\frac{1}{27}\cdot \frac{(1-2i)(2-i)}{5}=
\frac{1}{27} \frac{2-i-4i+2i^2}{5}=
\frac{1}{27} \frac{-5i}{5}=
\frac{-i}{27}\\
z^3=\frac{i}{27}\\
...}\)

POZDRO

[profesorq]

czyli jakie to sa wsyzstkie pierwiastki??

[natkoza]

\(\displaystyle{ (z^2+13)(x^3-\frac{27-54i}{2+i} )=0 z^2+13=0 x^3-\frac{27-54i}{2+i}=0\Leftrightarrow z^2=-13\vee z^3-\frac{(27-54i)(2-i)}{(2+i)(2-i)}=0\Leftrightarrow z^2=-13\vee z^3-\frac{54-504i-27i+54}{3}=0 z^3-\frac{-477i}{3}=0 z^2=-13 z^3-159i=0}\)
Rozwiażmu zatem kazde z równań osobno:
\(\displaystyle{ z^2=-13\\
z=a+bi\\
(a+bi)^2=-13 a^2+2abi-b^2=-13 \\
\begin{cases} a^2-b^2=-13 \\ 2abi=0 \end{cases} \\
\begin{cases} a^2-b^2=-13 \\ ab=0 \end{cases} \\
\begin{cases} a^2-b^2=-13 \\ a=0 b=0 \end{cases} \\
\begin{cases} -b^2=-13 a^2=-13\\ a=0 b=0 \end{cases}\\
\begin{cases} b^2=13 \\ a=0 b=0 \end{cases}\\
\begin{cases} b=\sqrt{13} b=-\sqrt{13} \\ a=0 \end{cases}\\}\)
Zatem
\(\displaystyle{ z_1=\sqrt{13}i\\
z_2=-\sqrt{13}i}\)
\(\displaystyle{ z^3-159i=0\\
z=a+bi\\
(a+bi)^3=159i\Leftrightarrow a^3+3a^2bi-3ab^2-b^3i=159i \\
\begin{cases} a^3-3ab^2=0 \\ 3a^2bi-b^3i=159i \end{cases}\\}\)
Rozwiazując ten układ dostaniesz kolejne wartości a i b

[profesorq]

Drugi element:
\(\displaystyle{ \frac{27-54i}{2+i}=
\frac{(27-54i)(2-i)}{4-i^2}=
\frac{1}{27}\cdot \frac{(1-2i)(2-i)}{5}=
\frac{1}{27} \frac{2-i-4i+2i^2}{5}=
\frac{1}{27} \frac{-5i}{5}=
\frac{-i}{27}\\
z^3=\frac{i}{27}\\
...}\)

POZDRO

czy tu nie ma czasem błedu? NIe powinno być czasem:
\(\displaystyle{ \frac{27-54i}{2+i}=
\frac{(27-54i)(2-i)}{4-i^2}=
\frac{27(1-2i)(2-i)}{5}=
\frac{27(2-i-4i+2i^2)}{5}=
27 \frac{-5i}{5}=
-27i\\

z^3-(3i)^3=0\\
(z-3i)(z^2+3iz+9i^2)=0\\
z=3i \ \ \ \ (z^2+3iz+9i^2)=0\\
\delta= 9i^2-4 9i^2=27\\
z_1=\frac{3\sqrt{3}}{2}-\frac{3i}{2}\\
z_2=\frac{-3\sqrt{3}}{2}-\frac{3i}{2}\\
\\}\)

Wszystkie zespolone pierwiastki tego równania to:
\(\displaystyle{ z_1=\frac{3\sqrt{3}}{2}-\frac{3i}{2}\\
z_2=\frac{-3\sqrt{3}}{2}-\frac{3i}{2}\\
z_3=3i\\
z_4=\sqrt{13}i\\
z_5=-\sqrt{13}i\\}\)

dobrze to rozwiązałem??