the complex number Z satisfies
Z/(Z+4)= 3+2i
Find Z.
please help thx!
Multipy both sides of the equation by Z+4, and combine terms.
Z = (3 + 2i)(Z+4) = 3Z + 2iZ +12 + 8i
(2 + 2i)Z + 12 + 8i = 0
Z = -(12 + 8i)/(2 + 2i)
= -(6 + 4i)/(i + 1)
= -(6 + 4i)(i -1)/[(i+1)(i -1)]
= -(-6 +2i -4)/(-2)
Z = -5 + i
i don't really get how 3Z + 2iZ +12 + 8i can be transformed into (2 + 2i)Z + 12 + 8i = 0
why is it not (3 + 2i)Z + 12 + 8i = 0??
thank you
Compare the first two equations.
I subtracted Z from both sides of the first equation, to get a zero on one side of the equation.
thanks!
To find the complex number Z, we can start by multiplying both sides of the equation by (Z + 4):
Z/(Z + 4) = 3 + 2i
Z = (Z + 4)(3 + 2i)
Now, let's expand the right side of the equation by distributing:
Z = 3Z + 2iZ + 12 + 8i
Next, let's rearrange the equation by bringing all the Z terms to one side and all the constant terms to the other side:
Z - 3Z - 2iZ = 12 + 8i
Combining like terms:
-2Z - 2iZ = 12 + 8i
Now, let's factor out Z from the left side:
Z(-2 - 2i) = 12 + 8i
Divide both sides by (-2 - 2i):
Z = (12 + 8i)/(-2 - 2i)
To rationalize the denominator, we need to multiply both the numerator and the denominator by the conjugate of (-2 - 2i), which is (-2 + 2i):
Z = (12 + 8i)/(-2 - 2i) * (-2 + 2i)/(-2 + 2i)
Expanding the numerator and denominator:
Z = (-24 + 16i + 16i - 32i^2) / (4 - 4i - 4i + 4i^2)
Since i^2 = -1, simplifying further:
Z = (-24 + 32i + 32i + 32) / (4 + 4)
Combining like terms:
Z = (8 + 64i) / 8
Finally, dividing both the numerator and the denominator by 8:
Z = 1 + 8i
Therefore, the complex number Z that satisfies the equation Z/(Z + 4) = 3 + 2i is Z = 1 + 8i.