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Question Number 112751 by bemath last updated on 09/Sep/20

Answered by bobhans last updated on 09/Sep/20

let w = a+bi and z = p+qi where z^− =p−qi ⇒w−2z = a+bi−2p−2qi = 9 ⇒(a−2p)+(b−2q)i = 9+0.i → { ((a−2p=9→a=9+2p)),((b= 2q)) :} ⇒3w−wz^− =(3a+3bi)−(a+bi)(p−qi)=17−30i (3a+3bi)−(ap−aqi+bpi+bq)=17−30i 3a+3bi−(ap+bq+(bp−aq)i)=17−30i (3a−ap−bq)+(3b−bp+aq)i=17−30i → { ((3a−ap−bq=17)),((3b−bp+aq=−30)) :} → { ((27+6p−9p−2p^2 −2q^2 =17)),((6q−2pq+9q+2pq=−30)) :} → { ((27−3p−2p^2 −2q^2 =17)),((15q=−30, q=−2 then b=−4)) :} (∗)27−3p−2p^2 −8=17 2p^2 +3p−2=0, (2p−1)(p+2)=0 { ((p=(1/2) →a=9+1=10)),((p=−2→a=9−4=5)) :} therefore case(1) w=10−4i & z=(1/2)−2i case(2)w=5−4i & z=−2−2i

letw=a+biandz=p+qiwherez−=p−qi⇒w−2z=a+bi−2p−2qi=9⇒(a−2p)+(b−2q)i=9+0.i→{a−2p=9→a=9+2pb=2q⇒3w−wz−=(3a+3bi)−(a+bi)(p−qi)=17−30i(3a+3bi)−(ap−aqi+bpi+bq)=17−30i3a+3bi−(ap+bq+(bp−aq)i)=17−30i(3a−ap−bq)+(3b−bp+aq)i=17−30i→{3a−ap−bq=173b−bp+aq=−30→{27+6p−9p−2p2−2q2=176q−2pq+9q+2pq=−30→{27−3p−2p2−2q2=1715q=−30,q=−2thenb=−4(∗)27−3p−2p2−8=172p2+3p−2=0,(2p−1)(p+2)=0{p=12→a=9+1=10p=−2→a=9−4=5thereforecase(1)w=10−4i&z=12−2icase(2)w=5−4i&z=−2−2i

Commented by bemath last updated on 09/Sep/20

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