Welche Teilmenge der komplexen Zahlenebene beschreibt die angegebene Ungleichung? |z+4z−4|<3{\displaystyle \left|{\frac {z+4}{z-4}}\right|<3}
von --Christian.abila 11:38, 17. Jul. 2012 (CEST)
Re(z+4z−4)ϵ]3,3[{\displaystyle \operatorname {Re} \left({\frac {z+4}{z-4}}\right)\epsilon ]3,3[} Im(z+4z−4)ϵ]3,3[{\displaystyle \operatorname {Im} \left({\frac {z+4}{z-4}}\right)\epsilon ]3,3[} Anders ausgedrückt: alle komplexen Zahlen, die sich innerhalb des Kreises mit Radius 3 befinden. Re(z+4z−4)2+Im(z+4z−4)2<3|2{\displaystyle {\sqrt {\operatorname {Re} \left({\frac {z+4}{z-4}}\right)^{2}+\operatorname {Im} \left({\frac {z+4}{z-4}}\right)^{2}}}<3|^{2}} Re(z+4z−4)2+Im(z+4z−4)2<9{\displaystyle \operatorname {Re} \left({\frac {z+4}{z-4}}\right)^{2}+\operatorname {Im} \left({\frac {z+4}{z-4}}\right)^{2}<9} z=a+ib,z¯=a−ib{\displaystyle z=a+ib,{\overline {z}}=a-ib} z1=z+4=a+ib+4=4+a+ib,a1=4+a{\displaystyle z_{1}=z+4=a+ib+4=4+a+ib,a_{1}=4+a} z2=z−4=a+ib−4=−4+a+ib,a2=−4+a{\displaystyle z_{2}=z-4=a+ib-4=-4+a+ib,a_{2}=-4+a} z+4z−4=z1z2=z1∗z2¯z2∗z2¯=(a1+ib)(a2−ib)(a2+ib)(a2−ib)=a1a2−iba1+iba2+b2a22−iba2+iba2+b2=a1a2+b2+ib(a2−a1)a22+b2{\displaystyle {\frac {z+4}{z-4}}={\frac {z_{1}}{z_{2}}}={\frac {z_{1}*{\overline {z_{2}}}}{z_{2}*{\overline {z_{2}}}}}={\frac {(a_{1}+ib)(a_{2}-ib)}{(a_{2}+ib)(a_{2}-ib)}}={\frac {a_{1}a_{2}-iba_{1}+iba_{2}+b^{2}}{a_{2}^{2}-iba_{2}+iba_{2}+b^{2}}}={\frac {a_{1}a_{2}+b^{2}+ib(a_{2}-a_{1})}{a_{2}^{2}+b^{2}}}} Re(z1z2)=a1a2+b2a22+b2,Im(z1z2)=b(a2−a1)a22+b2{\displaystyle \operatorname {Re} \left({\frac {z_{1}}{z_{2}}}\right)={\frac {a_{1}a_{2}+b^{2}}{a_{2}^{2}+b^{2}}},\operatorname {Im} \left({\frac {z_{1}}{z_{2}}}\right)={\frac {b(a_{2}-a_{1})}{a_{2}^{2}+b^{2}}}} (a1a2+b2a22+b2)2+(b(a2−a1)a22+b2)2<9{\displaystyle \left({\frac {a_{1}a_{2}+b^{2}}{a_{2}^{2}+b^{2}}}\right)^{2}+\left({\frac {b(a_{2}-a_{1})}{a_{2}^{2}+b^{2}}}\right)^{2}<9}