+0  
 
0
38
1
avatar

Describe all solutions to zw - 3w - 2iw + 4z = 7iw + 5w + 11iz - 13 + 15i
where z and w are complex numbers.

 Dec 29, 2022
 #1
avatar+1317 
+2

zw - 3w - 2iw + 4z = 7iw + 5w + 11iz - 13 + 15i

z = a + bi, w = c + di

(a + bi)(c + di) - 3(c + di) - 2i(c + di) + 4(a + bi) = 7i(c + di) + 5(c + di) + 11i(a + bi) - 13 + 15i

ac - bd + (bc + ad)i - 3c - 3di - 2ci + 2d + 4a + 4bi = 7ci - 7d + 5c + 5di + 11ai - 11b - 13 + 15i

Combine like terms.

[ac - bd - 3c + 2d + 4a] + [bci + adi - 3di - 2ci + 4bi] = [-7d + 5c - 11b - 13] + [7ci + 5di + 11ai + 15i]

Set Re = Re2, and Im = Im2

ac - bd - 3c + 2d + 4a = -7d + 5c - 11b - 13

ac - bd + 4a + 11b - 8c + 9d = -13

 

bci + adi - 3di - 2ci + 4bi = 7ci + 5di + 11ai + 15i

bc + ad - 3d - 2c + 4b = 7c + 5d + 11a + 15

ad + bc - 11a + 4b - 9c - 8d = 15

Notice how many of the coefficients seem swapped. This might imply something about conjugates. GL (you do it now)

 Dec 30, 2022

21 Online Users

avatar