Asked by Jude
Find the real values x and y given that 12+3i/x-iy=-1-i
Answers
Answered by
mathhelper
assuming you mean:
(12+3i)/(x-iy)=-1-i , those brackets are critical!
(x - iy)(-1 - i) = 12 + 3i
-x - xi + iy + i^2 y = 12 + 3i
-x - xi + iy - y = 12 + 3i
matching up real and complex
<b>-x - y = 12</b> and
-xi + iy = 3i or <b>x - y = -3</b>
add the two bold equations:
-2y = 9
y = -9/2
then x = -15/2
(12+3i)/(x-iy)=-1-i , those brackets are critical!
(x - iy)(-1 - i) = 12 + 3i
-x - xi + iy + i^2 y = 12 + 3i
-x - xi + iy - y = 12 + 3i
matching up real and complex
<b>-x - y = 12</b> and
-xi + iy = 3i or <b>x - y = -3</b>
add the two bold equations:
-2y = 9
y = -9/2
then x = -15/2