Równanie wielomianowe.

[BeHappy]

\(\displaystyle{ z^{2}+(3-2j)z-6j=0}\)

\(\displaystyle{ \Delta=9-6j-6j-4+24j=5+12j}\)

\(\displaystyle{ \sqrt{\Delta}=\sqrt{5+12j}}\)

\(\displaystyle{ \begin{cases} x^{2}-y^{2}=5 \\ 2xy=12 \\ x^{2}+y^{2}=13 \end{cases}}\)

\(\displaystyle{ 2x^{2}=18}\)

\(\displaystyle{ \sqrt{\Delta}=\sqrt{5+12j}= \begin{cases} 3+2j \\ -3-2j \end{cases}}\)

\(\displaystyle{ z_{1}=\frac{-3+2j+3+2j}{2}=2j}\)

\(\displaystyle{ z_{2}=\frac{-3+2j-3-2j}{2}=-3}\)


Dobrze?

[a4karo]

dobrze