Question

Sa se determine z€C astfel încât :
a) |z|=|z+1|=|z^-2|
b) |z+i|=|z+1| = |conjugat de z + 3i|
c) |z-1| +2z=17+9i

Answer 1

a)
Din |z|=|z^(-2)| obtinem |z|=1/(|z|^2) <=> |z|^3=1 <=> |z|=1.
Deci |z+1|=1.
Notam z=a+bi, unde |z|^2=a^2+b^2=1.
Obtinem |z+1|=1 <=> |(a+1)+bi|=1 <=> rad((a+1)^2+b^2)=1 <=> a^2+b^2+2a+1=1 <=> 2a+2=1 <=> 2a=-1 <=> a=-1/2. => b=+-rad(3)/2.

Rezulta z1=-1/2+i*rad(3)/2 si z2=-1/2-i*rad(3)/2. (ambele convin - se verifica inlocuind in relatiile initiale)

b)
analog cu a)

c)
Notam z=a+bi.
Atunci
|z-1|+2z=17+9i <=> |(a-1)+bi|=(17-2a)+(9-2b)i.
Cum |(a-1)+bi| apartine R => (17-2a)+(9-2b)i apartine R. => 9-2b=0 <=> b=9/2.

Obtinem deci |(a-1)+i*9/2|=17-2a <=> rad((a-1)^2+81/4)=17-2a => (a-1)^2+81/4=(17-2a)^2 <=> a^2-2a+85/4=289-68a+4a^2 <=> 3a^2-66a+1071/4=0 <=> a^2-22a+357/4=0
delta=(-22)^2-4*1*357/4=484-357=127.
a1,2=(68+-rad(127))/2.

Deci z1=(68+rad(127))/2+i*9/2 si z2=(68-rad(127))/2+i*9/2.