In my exercise book I couldn't find a way to solve what I've typed below (I know for final solutions anyway):
Find Re(w) and Im(w) [where ]:
So we have to write a complex number in a standard form and then extract a real and an imaginary part from there. We work here with the unknown variables ( and ).
My way would be:
If I expand the expression further (i.e. numerator) I don't find a way to write it in a standard form.
However, the final solution is:
Does anybody know the steps leading to the final solution?
Find all solutions to the equation [where ]:
It is a quadratic equation and has a conjugate of complex number in it.
Thanks for help in advance.
Anyway, I tried the 1st Problem only. I did expand it paiently and I came up to the shown Final solution.
W = [((x +yi)^2 +1)(x -yi)] / (x^2 +y^2)
W = [(x^2 +2xyi -y^2 +1)(x -yi)] / (x^2 +y^2)
W = [x^3 -(x^2)yi +2(x^2)yi -2x(y^2)(i^2) -xy^2 +(y^3)i +x -yi] / (x^2 +y^2)
W = [x^3 -(x^2)yi +2(x^2)yi +2xy^2 -xy^2 +(y^3)i +x -yi] / (x^2 +y^2)
W = [x^3 +(x^2)yi +xy^2 +(y^3)i +x -yi] / (x^2 +y^2)
Separate the imaginaries, or collect the reals and the imaginaries separately
W = [(x^3 +xy^2 +x) +i(yx^2 +y^3 -y)] / (x^2 +y^2)
Re(W) = (x^3 +xy^2 +x) /(x^2 +y^2)
Im(W) = [(y^3 +yx^2 -y) /(y^2 +x^2)](i).
Do long divisions on both and you'd arrive to the shown solution.
2(x + iy)^2 - 3(x - iy)^2 = 10i
2x^2 + 4ixy - 2y^2 - 3x^2 + 6ixy + 3y^2 = 10i
(-x^2 + y^2 ) + i(10xy - 10) = 0
Now, both the real and imaginary parts of this expression must be 0 so we have the system of equations:
-x^2 + y^2 = 0
10xy - 10 = 0
I would solve the bottom equation for y:
10xy = 10
xy = 1
y = 1/x
and insert this into the top equation:
-x^2 + (1/x)^2 = 0
-x^4 + 1 = 0
x^4 = 1
Now, remember that x is a REAL number. There are only two real solutions to this equation:
x = -1, 1
So y = 1/x means that y = -1, 1 respectively.
Thus your solution set is (-1, -1), (1, 1). Thus z = -1 - i and z = 1 + i. You can check these yourself.