Question

Complex Equation
|x + iy| = y - ix

Answer 1

ok so i have most of it but am not exactly sure about 1 part
this is what i have so far

Answer 2

z = x + iy
\[\left| z \right|= \sqrt{x^2 + y^2}\]

Answer 3

so \[\sqrt(x^2 + y^2) = y - ix\]

Answer 4

which comes out to x = -i

Answer 5

so x = 0

Answer 6

then the y part i'm not sure about... do you plug x=0 back into the original equation? if so, it comes out to y = |0| and the back of the book says the answer is \[y \ge 0\]

Answer 7

I want to suggest you some thing....

given, |x + iy| = y - ix
first, apply square root on both sides....

Answer 8

oh shoot... it actually comes out to |iy| = y so then |iy| = 0

Answer 9

@chetan552 yah, sqrt with the x and iy squared

Answer 10

er just y squared but yah

Answer 11

I don't know to what extend i am correct but i will try my level best...
(x + iy)^2 = (y - ix)^2

x^2+(iy)^2+2ixy=y^2+(ix)^2-2ixy

hear, we know that i^2=-1
replace it....

Answer 12

naw you can use euler's transformation. you are close but it would be
sqrt( x^2 + y^2 ) = y - ix
then as you did, you square both sides which gives u
x^2 + y^2 = y^2 -2ixy -x^2

Answer 13

both y^2 go away and you solve giving x = -i
then by doing a sort of equating coefficients sort of thing, the x is a real number on the left and there is no real number on the right so x = 0

Answer 14

that part is for sure right. it is the y part that i'm unsure about

Answer 15

yes that's it.....

Answer 16

no the solution in the solution manual says x = 0, y >= 0

Answer 17

i got it down to |iy| = 0 but why is that y=0? tho i prob did that part wrong

Answer 18

Oh then i am sorry...;)

Answer 19

=/ thanks for the help tho