Question

Numbers z satisfying the equation (3 + 7i)z + (10 2i)z + 100 = 0 represents

Answer 1

Let z = a+bi where a and b are real numbers.

(3 + 7i)z + (10 - 2i)z(bar) + 100 = 0

(3+7i)(a+bi) + (10-2i)(a-bi) + 100 = 0

(3a-7b) + i(7a+3b) + (10a-2b) - i(2a+10b) + 100 = 0

(13a - 9b + 100) + i(5a-7b) = 0

The real parts of both sides must equal: 13a-9b+100=0

The imaginary parts of both sides must equal: 5a-7b=0

Clearly this linear system of equations only has one solution. Thus there is only one value of z satisfying (3 + 7i)z + (10 - 2i)z(bar) + 100 = 0.