równanie l.zespolone

[profesorq]

1.Rozwiazac równanie, przedstawic interpretacje
graficzna rozwiazania.
\(\displaystyle{ (|8+6i| \frac{-1+2i}{2-i})^2=z^4}\)


2.jak Zaznaczyc na płaszczyznie zespolonej zbiór
\(\displaystyle{ \{z C : |z+7Re(\frac{\sqrt{3}}{2}- \frac{1}{2}i)^{84}|=|z+\frac{3-7i}{1-i}|\}}\)

[Rogal]

Jakie problemy?

[profesorq]

2)
\(\displaystyle{ \{z C : |z+7Re(\frac{\sqrt{3}}{2}- \frac{1}{2}i)^{84}|=|z+\frac{3-7i}{1-i}|\}}\)

\(\displaystyle{ *(\frac{\sqrt{3}}{2}- \frac{1}{2}i)^{84}=(cos\frac{\pi}{6}+isin\frac{\pi}{6})^{84}=cos14\pi +isin14\pi)*}\)

\(\displaystyle{ |z+7|=|z+\frac{3-7i}{1-i}|\}}\)
\(\displaystyle{ |z+7|=|z+5-2i|}\)
\(\displaystyle{ |x+iy+7|=|x+iy+5-2i|}\)
\(\displaystyle{ (x+7)^2+y^2=(x+5)^2+(y-2)^2}\)
\(\displaystyle{ 4x+4y-20=0}\)
\(\displaystyle{ y=-x+5}\)

dobrze?

[ Dodano: 28 Stycznia 2008, 22:03 ]
1)
\(\displaystyle{ (|8+6i| \frac{-1+2i}{2-i})^2=z^4}\)
\(\displaystyle{ (|8+6i| \frac{-1+2i}{2-i})=+/-z^2}\)
\(\displaystyle{ (10 \frac{(-1+2i)(2+i)}{(2-i)(2+i)})=+/-z^2}\)
\(\displaystyle{ 10 \frac{(4i-2-2-i)}{5}=+/-z^2}\)
\(\displaystyle{ 6i-8=z^2}\) lub \(\displaystyle{ -6i+8=z^2}\)
1)
\(\displaystyle{ z=a+bi}\)
\(\displaystyle{ (a+bi)^2=6i-8}\)
\(\displaystyle{ a^2+2abi-b^2=6i-8}\)
\(\displaystyle{ a^2-b^2=-8}\) i \(\displaystyle{ 2ab=6}\)
czyli \(\displaystyle{ a=\frac{3}{b}}\)
\(\displaystyle{ \frac{9}{b^2}-b^2=-8}\)
\(\displaystyle{ b^4-8b^2-9=0}\)
\(\displaystyle{ \delta = 100}\)
\(\displaystyle{ (I)b^2=-1}\) odrzucam
\(\displaystyle{ (II)b^2=9}\)

\(\displaystyle{ b_1=3}\) lub \(\displaystyle{ b_2=-3}\)
\(\displaystyle{ a_1=1}\) \(\displaystyle{ a_2=-1}\)

\(\displaystyle{ z_0=1+3i}\)
\(\displaystyle{ z_1=-1-3i}\)

2)
\(\displaystyle{ z=a+bi}\)
\(\displaystyle{ (a+bi)^2=-6i+8}\)
\(\displaystyle{ a^2+2abi-b^2=-6i+8}\)
\(\displaystyle{ a^2-b^2=8}\) i \(\displaystyle{ 2ab=-6}\)
czyli \(\displaystyle{ a=-\frac{3}{b}}\)
\(\displaystyle{ \frac{9}{b^2}-b^2=8}\)
\(\displaystyle{ b^4+8b^2-9=0}\)
\(\displaystyle{ \delta = 100}\)
\(\displaystyle{ (I)b^2=1}\)
\(\displaystyle{ (II)b^2=-9}\) odrzucam

\(\displaystyle{ b_1=1}\) lub \(\displaystyle{ b_2=-1}\)
\(\displaystyle{ a_1=-3}\) \(\displaystyle{ a_2=3}\)

\(\displaystyle{ z_2=-3+i}\)
\(\displaystyle{ z_3=3-i}\)
[ Dodano: 28 Stycznia 2008, 22:07 ]
dobrze?