Stuck on imaginary number simultaneous equation

Flyingpooman

New member
Joined
Oct 11, 2016
Messages
1
Solve this simultaneous equation:

z^2 - w^2 = -2+16i

z + iw = 4+4i

please help I tried to do this for so long and am struggling. I tried substituting z as well as w from the second equation into the first one but I just cannot end up with the values of z and w which are in the form of a+bi.
 
Solve this simultaneous equation:

z^2 - w^2 = -2+16i

z + iw = 4+4i

please help I tried to do this for so long and am struggling. I tried substituting z as well as w from the second equation into the first one but I just cannot end up with the values of z and w which are in the form of a+bi.

[z + iw]2 = [4+4i]2

continue.......
 
Multiply [2] by 4, and rearrange:
z^2 - w^2 + 2 = 16i [1]
4z + 4iw - 16 = 16i [2]

Subtract to get:
z^2 - 4z - w^2 - 4iw + 18 = 0

You now have a quadratic...carry on...

Denis, but you have one quadratic equation in two variables.

How can z and w be isolated?
 
Solve this simultaneous equation:

z^2 - w^2 = -2+16i

z + iw = 4+4i

please help I tried to do this for so long and am struggling. I tried substituting z as well as w from the second equation into the first one but I just cannot end up with the values of z and w which are in the form of a+bi.
Assuming we are looking for values of z and w and that i is the 'positive' square root of -1, there is no solution IF both z and w are real as the equations are written. Thus either (or both) z and w must be complex numbers. So letting
z = zr + i zi
and
w = wr + i wi
where zr, zi, wr, and wi are real valued. The second equation gives
z = a + w = 4 + wr + i (4 + wi)
where a = 4 + 4 i. Equating real and imaginary parts, the first equation gives
(a) zr2-zi2-wr2+wi2 = -2
(b) zr zi + wr wi = 8
See if you can go from there.

Hint: Start with (a) to get a linear equation in wr and wi
 
Last edited:
Top