Question

Solve the pair of simultaneous equations for the complex numbers z and w.

(1+i)z + (2-i)w=3+4i
iz + (3+i)w=-1+5i

Answer 1

(z+2w)+i(z-2w) = 2 + 4i =>z+2w=2, z-2w=4

Answer 2

from second w=5, z+3w=-1 => w=-2

Answer 3

z=2, w=i are the answers...? How to get there?

Answer 4

For (1+i)z + (2-i)w=3+4i
(1+i)z + (2-i)w=3+4i
z +zi + 2w - wi = 3+4i
(z+2w) + (z-w) i = 3+4i
So you get
z+2w =1
z-w =4

Answer 5

z+2w =1 why?

Answer 6

Like the previous one, compare the real part and the imaginary part to get the 2 equations

Answer 7

Sorry. that should be 3

Answer 8

oops sorry .. i thought w and z were real numbers

Answer 9

z+2w =3

Answer 10

So, how do you get w as i in the final answer?

Answer 11

Wait ... i thought i read that wrong

Answer 12

let z = a+ib and w = c+id

Answer 13

(1+i)z + (2-i)w=3+4i
=>(1+i)( a+ib) + (2-i)(c+id)=3+4i
=> (a-b)+(a+b)i+(2c+d)+(2d-c)i = 3+4i
=> (a-b+2c+d)+i(a+b+2d-c) = 3+4i
=> (a-b+2c+d) = 2
and
=>(a+b+2d-c) = 4

Answer 14

for other part

i( a+ib) + (3+i)(c+id)=-1+5i
=> -b+ai+(3c-d)+(3d+c)i = -1+5i
=> (-b+3c-d)+i(a+3d+c) = -1 + 5i
=> (-b+3c-d) = -1
and
=> (a+3d+c) = 5

Answer 15

4 unknowns 4 equations .. pretty messy let's ask wolf for answer

Answer 16

there you have your 4 quantities
http://www.wolframalpha.com/input/?i=+%28a-b%2B2c%2Bd%29+%3D+2%3B+%28a%2Bb%2B2d-c%29+%3D+4%3B+%28-b%2B3c-d%29+%3D+-1%3B++%28a%2B3d%2Bc%29+%3D+5
put the values in w and z and you have your answer

Answer 17

I got it!!! :D

Answer 18

\[(1+i)z+(2-i)w=3+4i\]\[iz+(3+i)w=-1+5i\]
\[z={(-1+5i)-(3+1)w \over i}\]
\[(1+i){(-1+5i)-(3+i)w \over i}+(2w-iw)=3+4i\]

---> which simplifies as, after lots of calculation --->
\[w=-3+10i+3 ----> w=i\]
(There are 20 steps involved in getting the answer...)

Answer 19

@Callisto @experimentX

Answer 20

yeah ... a better way!!