simultaneous with complex (1 Viewer)

[CriminalCrab]

Solve for z and w:
1) (2+i)z+(2-i)w=1 and (2-i)z+(2+i)w=2

[ans: z=(3/8)+(i/4) w=(3/8)-(i/4) ]

2) z+(1-i)w=2i and w+(1-i)z=1

[ans: z=1+i w=-1]

thanks

 

[RealiseNothing]

I'll do the 'z' for the first one, you can then do the rest.

Add together the two equations:

z(2+i) + z(2-i) + w(2-i) + w(2+i) = 1 + 2

z(2+i+2-i) + w(2-i+2+i) = 3

4z + 4w = 3

Re-arrange, make 'w' the subject:

4w = 3 - 4z

w = (3-4z)/4

Substitute this value into one of the equations:

z(2+i) + [(2-i)(3-4z)]/4 = 1

Get rid of fraction:

4z(2+i) + (2-i)(3-4z) = 4

8z + 4iz + 6 - 3i - 8z + 4iz = 4

8iz - 3i + 6 = 4

8iz = 3i - 2

z = (3i-2)/8i

Split the fraction:

z = 3i/8i - 2/8i

z= 3/8 - 1/4i

Now just do the same thing but find the value of 'z' in terms of 'w' this time and substitute that in.