Równanie zespolone

[kasiaak]

Witam, mam problem... nie mam pojęcia o rozwiązywaniu równań zespolonych, a potrzebuje je ogarnąć na zaliczenie... Oto kilka przykładów, prosiłabym o rozwiązanie:

1) \(z^2-5z+7+i=0\)
2) \((1-i)z+2i=0\)
3) \(z^2-(2-i)z-2i=0\)
4)\(z^2-(2-i)z+1-i=0\)
5) \(iz^2-2iz+2+i=0\)
6) \(z^3=i\)
7) \(z^3+4z^2+4z+1=0\)
8 ) \(z^4-z^3-z+1=0\)
9) \(z^3(z^2-z+1)=0\)
10) \((z^4+16)(z^2+z+4)=0\)


Nie dopisuj swoich zadań do cudzego tematu. Zakładaj własny

[irena]

1.
\(z^2-5z+7+i=0\\\Delta=25-28-4i=-3-4i\\(a+bi)^2=a^2-b^2+2abi=-3-4i\\\{a^2-b^2=-3\\2ab=-4\)

\(b=-\frac{2}{a}\\a^2-\frac{4}{a^2}=-3\ /\cdot a^2\\a^4+3a^2-4=0\\a^2=1\ \vee\ a^2=-4\\a\in R\\a^2=1\\a=1\ \vee\ a=-1\)

\(\{a=1\\b=-2\) lub \(\{a=-1\\b=2\)

\(\sqrt{\Delta}=1-2i\ \ \vee\ \ \sqrt{\Delta}=-1+2i\)

\(z_1=\frac{5-1+2i}{2}=-2+i\ \ \vee\ \ z_2=\frac{5+1-2i}{2}=3-i\)

[irena]

2.
\((1-i)z+2i=0\\(1-i)z=-2i\ /\cdot(1+i)\\(1-i^2)z=-2i-2i^2\\(1+1)z=-2i+2\\z=1-i\)

[irena]

3.
\(z^2-(2-i)z-2i=0\\\Delta=(2-i)^2+8i=4-4i+i^2+8i=3+4i\\(a+bi)^2=3+4i\\a^2-b^2+2abi=3+4i\\\{a^2-b^2=3\\2ab=4\)

\(b=\frac{2}{a}\\a^2-\frac{4}{a^2}=3\ /\cdot a^2\\a^4-3a^2-4=0\\a^2=-1\ \vee\ a^2=4\\a\in R\\a=2\ \vee\ a=-2\)

\(\{a=2\\b=1\) lub \(\{a=-2\\b=-1\)

\(\sqrt{\Delta}=2+i\ \ \vee\ \ \sqrt{\Delta}=-2-i\)

\(z_1=\frac{2-i-2-i}{2}=-i\ \ \vee\ \ z_2=\frac{2-i+2+i}{2}=2\)

[irena]

4.
\(z^2-(2-i)z+1-i=0\\\Delta=(2-i)^2-4+4i=4-4i+i^2-4+4i=-1\\\sqrt{\Delta}=i\ \vee\ \sqrt{\Delta}=-i\\z_1=\frac{2-i-i}{2}=1-i\ \vee\ z_2=\frac{2-i+i}{2}=1\)

[irena]

5.
\(iz^2-2iz+2+i=0\ /\cdot(-i)\\-i^2z^2+2i^2z-2i-i^2=0\\z^2-2z-2i+1=0\\\Delta=4+8i-4=8i\\\{a^2-b^2=0\\2ab=8\)

\(a=b\ \vee\ a=ib\\2a^2=8\ \vee\ -2a^2=8\\a^2=4\ \vee\ a^2=-4\\a\in R\\a=b=2\ \vee\ a=b=-2\\\sqrt{\Delta}=2+2i\ \vee\ \sqrt{\Delta}=-2-2i\\z_1=\frac{2-2-2i}{2}=-i\ \vee\ z_2=\frac{2+2+2i}{2}=2+i\)

[irena]

6.
\(z^3=i\\i=0+i=cos(\frac{\pi}{2})+i sin(\frac{\pi}{2})\\z_1=cos(\frac{\pi}{6})+i sin(\frac{\pi}{6})=\frac{\sqrt{3}}{2}+\frac{1}{2}i\\z_2=cos(\frac{5}{6}\pi)+i sin(\frac{5}{6}\pi)=-\frac{\sqrt{3}}{2}+\frac{1}{2}i\\z_3=cos(\frac{3}{2}\pi)+i sin(\frac{3}{2}\pi)=0-i=-i\)

[irena]

7.
\(z^3+4z^2+4z+1=0\\(z+1)(z^2+3z+1)+0\\z+1=0\ \vee\ z^2+3z+1=0\\\Delta=9-4=5\\z_1=-1\ \vee\ z_2=\frac{-3-\sqrt{5}}{2}\ \vee\ z_3=\frac{-3+\sqrt{5}}{2}\)

[irena]

8.
\(z^4-z^3-z+1=0\\z^3(z-1)-(z-1)=0\\(z-1)(z^3-1)=0\\(z-1)(z-1)(z^2+z+1)=0\\(z-1)^2(z^2+z+1)=0\\(z-1)^2=0\ \vee\ z^2+z+1=0\\z-1=0\\z_1=1\\\Delta=-3\\\sqrt{\Delta}=i\sqrt{3}\\z_1=1\ \vee\ z_2=\frac{-1-i\sqrt{3}}{2}\ \vee\ z_3=\frac{-1+i\sqrt{3}}{2}\)

[irena]

9.
\(z^3(z^2-z+1)=0\\z^3=0\ \vee\ z^2-z+1=0\\z_1=0\\\Delta=1-4=-3\\\sqrt{\Delta}=i\sqrt{3}\\z_1=0\ \vee\ z_2=\frac{1-i\sqrt{3}}{2}\ \vee\ z_3=\frac{1+i\sqrt{3}}{2}\)

[irena]

10.
\((z^4+16)(z^2+z+4)=0\\z^4=-16\\-16=16(-1+0i)=16(cos\pi+i sin\pi)\\z_1=2(cos(\frac{\pi}{4})+i sin(\frac{\pi}{4}))=\sqrt{2}+\sqrt{2}i\\z_2=2(cos(\frac{3}{4}\pi)+i sin(\frac{3}{4}\pi))=-\sqrt{2}+i\sqrt{2}\\z_3=2(cos(\frac{5}{4}\pi)+i sin(\frac{5}{4}\pi))=-\sqrt{2}-i\sqrt{2}\\z_4=2(cos(\frac{7}{4}\pi)+i sin(\frac{7}{4}\pi))=\sqrt{2}-i\sqrt{2}\)

\(z^2+z+4=0\\\Delta=1-16=-15\\\sqrt{\Delta}=i\sqrt{15}\\z_5=\frac{-1-i\sqrt{15}}{2}\ \vee\ z_6=\frac{-1+i\sqrt{15}}{2}\)