Complex number & imaginary number problems ?

1) i [3(a - bi) + i(a + bi)] = (a + bi) + 1
==> i [(3a - 3bi) + (ai - b)] = (a + 1) + bi
==> (3ai + 3b) + (-a - bi) = (a + 1) + bi
==> (3b - a) + (3a - b) i = (a + 1) + bi

Equating real and imaginary parts:
3b - a = a + 1
3a - b = b.

Rewrite these as
3b - 2a = 1
a = 2b/3.

Substitute the second into the first:
3b - 2(2b/3) = 1
==> 5b/3 = 1
==> b = 3/5.
Thus, a = (2/3)(3/5) = 2/5.

Therefore, z = a + bi = 2/5 + 3i/5.
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2) z/(z + w)
= (a + bi) / [(a + b) + (b + a)i]
= (a + bi) / [(a + b)(1 + i)], by factoring
= [(a + bi)(1 - i)] / [(a + b)(1 + i)(1 - i)], multiplying by the conjugate (1 - i)
= [(a + b) + i(a - b)] / [(a + b) * 2].

Ignoring the imaginary part yields
Re(z/(z + w)) = (a + b) / [(a + b) * 2] = 1/2.

I hope this helps!