Szczegółowe wyjaśnienie:
[tex]a)\ z^3-6z^2+34z=0\\\\z(z^2-6z+34)=0\iff z=0\ \vee\ z^2-6z+34=0\\\\z^2-6z+34=0\\\\\Delta=(-6)^2-4\cdot1\cdot34=36-136=-100\\\sqrt\Delta=\sqrt{-100}=10i\\\\z_1=\dfrac{-(-6)-10i}{2\cdot1}=3-5i\\\\z_2=\dfrac{-(-6)+10i}{2\cdot1}=3+5i[/tex]
[tex]\dfrac{1}{z_1}+\dfrac{1}{z_2}=\dfrac{z_2+z_1}{z_1z_2}\\\\\dfrac{3+5i+3-5i}{(3-5i)(3+5i)}=\dfrac{6}{3^2-(5i)^2}=\dfrac{6}{9+25}=\dfrac{6}{34}=\dfrac{3}{17}[/tex]
[tex]b)\ z^4+6z^2+8=0\\\\z^2=t\\\\t^2+6t+8=0\\t^2+2t+4t+8=0\\t(t+2)+4(t+2)=0\\(t+2)(t+4)=0\iff t+2=0\ \vee\ t+4=0\\t=-2\ \vee\ t=-4\\z^2=-2\ \vee\ z^2=-4\\z=\pm\sqrt{-2}\ \vee\ z=\pm\sqrt{-4}\\z=-i\sqrt2\ \vee\ z=i\sqrt2 \vee\ z=-2i\ \vee\ z=2i[/tex]
[tex]\dfrac{1}{z_1}+\dfrac{1}{z_2}+\dfrac{1}{z_3}+\dfrac{1}{z_4}=\dfrac{z_2z_3z_4+z_1z_3z_4+z_1z_2z_4+z_1z_2z_3}{z_1z_2z_3z_4}\\\\\dfrac{(i\sqrt2)(-2i)(2i)+(-i\sqrt2)(-2i)(2i)+(-i\sqrt2)(i\sqrt2)(2i)+(-i\sqrt2)(i\sqrt2)(-2i)}{(-i\sqrt2)(i\sqrt2)(-2i)(2i)}\\\\=\dfrac{4i\sqrt2-4i\sqrt2+4i-4i}{8}=\dfrac{0}{8}=0[/tex]
[tex]c)\ z^2+15+8i=0\qquad|-15-8i\\z^2=-15-8i\\(x+yi)^2=-15-8i\\x^2+2ixy-y^2=-15-8i\Rightarrow\left\{\begin{array}{ccc}x^2-y^2=-15\\2xy=-8\end{array}\right\\\\\left\{\begin{array}{ccc}x^2-y^2=-15\\x=-\dfrac{4}{y}\end{array}\right\\\\\left(-\dfrac{4}{y}\right)^2-y^2=-15\\\\\dfrac{16}{y^2}-\dfrac{y^4}{y^2}+\dfrac{15y^2}{y^2}=0\\\\\dfrac{-y^4+15y^2+16}{y^2}=0\iff- y^4+15y^2+16=0\\\\t=y^2\geq0\\\\-t^2+15t+16=0\\\\\Delta_t=15^2-4\cdot(-1)\cdot16=225+64=289\\\sqrt\Delta_t=\sqrt{289}=17[/tex]
[tex]t_1=\dfrac{-15-17}{2\cdot(-1)}=\dfrac{-32}{-2}=16>0\\\\\\t_2=\dfrac{-15+17}{2\cdot(-1)}=\dfrac{2}{-2}=-1<0\\\\y^2=16\to y=\pm\sqrt{16}\\\\y=-4\ \vee\ y=4[/tex]
[tex]x=-\dfrac{4}{y}\to x_1=-\dfrac{4}{-4}=1\ \vee\ x_2=-\dfrac{4}{4}=-1[/tex]
[tex]z_1=1-4i\ \vee\ z_2=-1+4i[/tex]
[tex]\dfrac{1}{z_1}+\dfrac{1}{z_2}=\dfrac{z_2+z_1}{z_1z_2}\\\\\dfrac{-1+4i+1-4i}{(-1+4i)(1-4i)}=\dfrac{0}{(-1+4i)(1-4i)}=0[/tex]
[tex]d)\ z^3+27=0\\\\z^3+3^3=0\\\\(z+3)(z^2-3z+3^2)=0\iff z+3=0\ \vee\ z^2-3z+9=0\\\\z^2-3z+9=0\\\\\Delta=(-3)^2-4\cdot1\cdot9=9-36=-27\\\sqrt\Delta=\sqrt{-27}=3i\sqrt3\\\\z_1=\dfrac{-(-3)-3i\sqrt3}{2\cdot1}=\dfrac{3-3i\sqrt3}{2}=\dfrac{3}{2}-\dfrac{3i\sqrt3}{2}\\\\z_2=\dfrac{-(-3)+3i\sqrt3}{2\cdot1}=\dfrac{3+3i\sqrt3}{2}=\dfrac{3}{2}+\dfrac{3i\sqrt3}{2}[/tex]
[tex]\dfrac{1}{z_0}+\dfrac{1}{z_1}+\dfrac{1}{z_2}=\dfrac{z_1z_2+z_0z_2+z_0z_1}{z_0z_1z_2}\\\\\dfrac{\frac{3-3i\sqrt3}{2}\cdot\frac{3+3i\sqrt3}{2}+(-3)\cdot\frac{3+3i\sqrt3}{2}+(-3)\cdot\frac{3-3i\sqrt3}{2}}{-3\cdot\frac{3-3i\sqrt3}{2}\cdot\frac{3+3i\sqrt3}{2}}\\\\=\dfrac{\frac{3^2-(3i\sqrt3)^2}{2^2}+\frac{-9-9i\sqrt3}{2}+\frac{-9+9i\sqrt3}{2}}{-3\cdot\frac{3^2-(3i\sqrt3)^2}{2^2}}\\\\=\bigg(\dfrac{9+27}{4}+\dfrac{-18}{2}\bigg):\bigg(\dfrac{-27-81}{4}\bigg)=(9-9):(-27)=0[/tex]
