Question

Dado z=(2x²-14)+(y²+3y)i; calcule os números reais x e y de forma que: a) z=4+4i b) z=10i​

Answer 1

Resposta:

a)

z=4+4i  =(2x²-14)+(y²+3y)i

4=2x²-14 ==>2x²=16 ==> x²=8  ==>x²=±√8=±2√2

4=y²+3y

y²+3y-4=0

y'=[-3+√(9+16)]/2=(-3+5)/2=1

y''=[-3-√(9+16)]/2=(-3-5)/2=-4

b)

z=10i​=(2x²-14)+(y²+3y)i

2x²-14=0  ==>x²=±√7

10=y²+3y

y²+3y-10=0

y'=[-3+√(9+40)]/2 =(-3+7)/2 =2

y''=[-3-√(9+40)]/2 =(-3-7)/2 =-5