Thank you. After I tried again, I got the same answers as yours too.

Let z = a + bi. Then z* = a - bi

z* = w/(1 + i)
w = z*(1 + i)


Given : z² + zw - 2 = 0

Sub w = z*(1 + i),

z² + z(z*(1 + i)) - 2 = 0


Recall that z(z*) = a² + b². Sub z = a + bi,

(a + bi)² + (a²+b²)(1 + i) - 2 = 0

a² + 2abi - b² + a² + b² + (a²+b²)i - 2 = 0

2a² + (2ab + a² + b²)i = 2 + 0i
2a² + (a + b)²i = 2 + 0i

Compare coefficients of the real part,

2a² = 2
a² = 1
a = 1 or a = - 1

Compare coefficients of the imaginary part,

(a + b)² = 0
a + b = 0
b = - a

So b = - 1 or b = 1

Therefore,

z = 1 - i or z = -1 + i

Since w = z*(1 + i)

w = (1 + i)(1 + i) or w = (-1 - i)(1 + i)

w = 1² + 2i - 1 or w = -(1+i)(1+i)

w = 2i or w = -2i


Therefore ,

z = 1 - i, w = 2i or z = -1 + i, w = - 2i


Note that since the possible values of of z are negatives of each other, then it follows that the possible values of w are also negatives of each other since the z* would be negatives of each other as well.

Thank you. Is it ok if I make w the subject for the 1st equation, which is w=((2-z)^)2/z?

You mean to solve w first?

No. Instead of making w the subject of the formula for the 2nd equation, can I do it for the 1st equation?

Yes, you can. Why not?

Given : z² + zw - 2 = 0

zw = 2 - z²
w = (2 - z²)/z


Sub this into z* = w/(1 + i) ,

z* = (2-z²)/(z(1 + i))

zz*(1 + i) = 2 - z²

It's the same no matter which one you use to make w the subject first. If done correctly with no error, you'll get to the same result (compare this with my previous working to see that it's the same.

For normal simultaneous equations it's like that too, you can choose either.to start with. This question is just simultaneous equations with a imaginary twist

Oh I see. Which method would be less tedious to do?

Not much difference actually. I usually do what comes to mind first.

If I really have to pick one, the first one would be it as I used fewer steps, and I don't have to divide then multiply back again to get rid of the denominator, unlike the second one.

Depends on the question also. Exercise your discretion and choose which one is easiest for you. Sometimes what others easy might be tedious and long to you, vice versa

Typically for me, I avoid fractions as far as possible. For some reason linear expressions (as in, no fractions whatsoever) are more appealing to my eye.

In this case, solving for w first would probably be more tedious. I also feel that the question will be tedious without the substitution z = a + bi.

I doubt it can be done without the substitution. The question gave the conjugate. How else to express the conjugate other than letting z = a + bi first? The substitution is required to put them all in the same kind of terms

Thanks for all your explanations!