Find the complex numbers z and W such that z+w=4+3i and z-iw=3-2i
We can solve for z and w using the system of linear equations:
z + w = 4 + 3i (multiply by i)
z - iw = 3 - 2i
When we multiply the first equation by i, we get:
iz + iw = 4i - 3 (add to the second equation)
z - iw = 3 - 2i
Adding these two equations gives:
2z = 4i - 3 + 3 - 2i
2z = 1 + 2i
z = (1 + 2i)/2
z = 1/2 + i
Substituting this value of z into the first equation gives:
1/2 + i + w = 4 + 3i
w = 3/2 + 2i
So the complex numbers z and w that satisfy the given system of equations are:
z = 1/2 + i
w = 3/2 + 2i
Looks like the bot is WRONG AGAIN, its answers don't verify, mine do
z+w=4+3i ,
z-iw=3-2i
subtract:
w + iw = 1 + 5i
w(1 + i) = 1 + 5i
w = (1+5i)/(1 + i) * (1 - i)/(1 - i)
= (1 + 4i - 5i^2)/(1 - i^2) = (6 + 4i)/2 = 3 + 2i
sub this back into z+w=4+3i
z + 3 + 2i = 4 + 3i
z = 1 + i
check:
z + w = 1 + i + 3 + 2i = 4 + 3i, check!
z - iw = 1 + i - i(3 + 2i)
= 1 + i - 3i - 2i^2
= 3 - 2i , check!!