# Solve for z, and give your answer in the form a+bi. z/(4+i) = z+5+4i?

### 6 Answers

- sepiaLv 72 weeks ago
z/(4 + i) = z + 5 + 4i

z = (4 + i)z + (4 + i)(5 + 4i)

z(1 - (4 + i) = (4 + i)(5 + 4i)

z(-3 - i) = 16 + 21i

z = (16 + 21i)/(-3 - i)

(16 + 21i)/(-3 - i) = a + bi

so 16 + 21i = (-3 - i)(a + bi) = (-3a + b) + (-a - 3b)i,

so -3a + b =16 and a + 3b = -21,

which yields

a = -6.9 and b = -4.7

- Φ² = Φ+1Lv 72 weeks ago
z/(4+i) = z+5+4i

z = (4+i)(z + 5+4i)

z = (4+i)z + (4+i)(5+4i)

z(-3-i) = 16+21i

z = (16+21i)/(-3-i)

⦙ (see below)

z = -6.9 - 4.7i

Source(s): Aside: (16+21i)/(-3-i) = a+bi so 16+21i = (-3-i)(a+bi) = (-3a+b)+(-a-3b)i, so -3a+b=16 and a+3b=-21, which yields a = -6.9 and b = -4.7 - rotchmLv 72 weeks ago
Just like the algebra learned in early highschool. Isolate the variable (z). What do you get?

z/(4+i) = z+5+4i. Put "z" to the left:

z/(4+i) -z = 5+4i. factor out z:

z(1/(4+i) - 1) = 5 + 4i. Divide by (1/(4+i) - 1) :

z = (5+4i) / (1/(4+i) - 1)

Done!

Now suffice to "clean it up". Again, its just highschool algebra & using i² = -1. You can thus rewrite the RHS as a + bi. I leave you to try it out a little. If you can't after an hour or so, ask and we can further walk you through.

- How do you think about the answers? You can sign in to vote the answer.
- 2 weeks ago
z / (4 + i) = z + 5 + 4i

z * (4 - i) / (16 - i^2) = z + 5 + 4i

z * (4 - i) / (16 - (-1)) = z + 5 + 4i

z * (4 - i) / (16 + 1) = z + 5 + 4i

z * (4 - i) / 17 = z + 5 + 4i

z * (4 - i) = 17z + 85 + 68i

z * (4 - i) - 17z = 85 + 68i

z * (4 - 17 - i) = 85 + 68i

z * (-13 - i) = 85 + 68i

z = (85 + 68i) / (-13 - i)

z = -(85 + 68i) / (13 + i)

z = -(85 + 68i) * (13 - i) / (13^2 - i^2)

z = -(85 + 68i) * (13 - i) / (169 - (-1))

z = -17 * (5 + 4i) * (13 - i) / (169 + 1)

z = -17 * (5 + 4i) * (13 - i) / 170

z = -(5 + 4i) * (13 - i) / 10

z = -(65 - 6i + 52i - 4i^2) / 10

z = -(65 + 46i + 4) / 10

z = -(69 + 46i) / 10

z = -6.9 - 4.6 * i

- billrussell42Lv 72 weeks ago
z/(4+i) = z+5+4i

z = a+bi

(a+bi)/(4+i) = a+bi+5+4i

multiply by 4+i

(a+bi) = (a+bi+5+4i)(4+i)

a+bi = 4a+4bi+20+16i +ai+bi²+5i+4i²

a+bi = 4a+4bi+20+16i +ai–b+5i–4

a+bi = 4a+4bi+16+16i +ai–b+5i

a+bi = 4a+16–b+i(4b+21+a)

set real and imag parts equal

a = 4a+16–b

–3a = 16–b

a = –(16/3) + (b/3)

b = 4b+21+a

–3b = 21+a

b = –7 – (a/3)

two equations in two unknowns

a = –(16/3) + (b/3)

b = –7 – (a/3)

a = –(16/3) + ((–7 – (a/3))/3)

a = –(16/3) – (7/3) – (a/9)

a + (a/9)= –(23/3)

a(10/9)= –(23/3)

a = –(9/10)(23/3) = –69/10

b = –7 – (a/3)

b = –7 – ((–69/10)/3)

b = –7 + (23/10)

b = –(70/10) + (23/10)

b = –47/10