Răspuns:
Explicație pas cu pas:
[tex]1)~~z=\dfrac{(1-i)(\sqrt{3}+i) }{1+i}=\dfrac{(1-i)(1-i)(\sqrt{3}+i) }{(1-i)(1+i)}=\dfrac{(1^{2}-2i+i^{2})(\sqrt{3}+i)}{1^{2}-i^{2}}=\dfrac{(1-2i-1)(\sqrt{3}+i)}{1+1}=-i\sqrt{3}-i^{2}=1-i\sqrt{3}.\\a=Re(z)+(Im(z))^{2}=1+(-\sqrt{3})^{2}=1+3=4.[/tex]
deci, a²=4²=16.
[tex]z=\dfrac{2a+i}{1-i} =\dfrac{(2a+i)(1+i)}{(1-i)(1+i)}=\dfrac{2a-1+(2a+1)i}{1^{2}-i^{2}}= \dfrac{2a-1+(2a+1)i}{1+1}=\dfrac{2a-1}{2}+\dfrac{2a+1}{2}i[/tex]
z∈R, dacă Im(z)=0, ⇒2a+1=0, ⇒a=-1/2.
Ex3. a) z²=-16, z=±√(-16)=±√(16·(-1))=±√(16·i²)=±4i.
b) 2x²-2x+1=0, Δ=(-2)²-4·2·1=4-8=-4=4·(-1)=4i²
x1=(2-2i)/4=(1-i)2, x2=(1+i)/2.
c) fie x²=z, ⇒ z²-9z-36=0, Δ=81+144=225, ⇒ z1=(9-15)/2=-3, iar z2=(9+15)/2=12
Deci x²=-3, ⇒x²=3·(-1) ⇒x²=3·i², ⇒ x1,2=±i·√3
și x²=12, ⇒x²=4·3, ⇒x3,4=±2√3
Răspuns: S={±i·√3; ±2√3}
[tex]4)~Fie~z=\dfrac{8+i}{7-4i}= \dfrac{(8+i)(7+4i)}{(7-4i)(7+4i)}=\dfrac{56+32i+7i+4i^{2}}{7^{2}-16i^{2}} =\dfrac{52+39i}{49+16}=\dfrac{52}{65}+\dfrac{39}{65}i=\dfrac{4}{5}+\dfrac{3}{5}i \\Atunci~|z|=\sqrt{(\frac{4}{5})^{2}+(\frac{3}{5})^{2} }=1[/tex]
