Das komplexe Polynom f(z)=z3+9jz2−28z−32j besitzt die Nullstelle z1 = - 4j
Polynomdivision:
(z3+9jz2−28z−32j):(z+4j)=z2+5jz-8
-(z3+4jz2)
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5jz2−28z
-(5jz2-20*z) weil 20j2=-20
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-8z−32j
-(-8z-32j)
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0
z2+5jz-8=0 |+8
z2+5jz=8
(z+\( \frac{5}{2} \)j)^2=8+\( \frac{25}{4} \)j^2=8-\( \frac{25}{4} \)=\( \frac{7}{4} \)|\( \sqrt{} \)
1.)z+\( \frac{5}{2} \)j=\( \frac{1}{2} \)\( \sqrt{7} \)
z₂=-\( \frac{5}{2} \)j+\( \frac{1}{2} \)\( \sqrt{7} \)
2.)z+\( \frac{5}{2} \)j=-\( \frac{1}{2} \)\( \sqrt{7} \)
z₃=-\( \frac{5}{2} \)j-\( \frac{1}{2} \)\( \sqrt{7} \)