Solving A Locus Problem | Problem 279

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[@chaosredefined3834]

|z / (z + i)| = 1 |z| / |z + i| = 1 |z| = |z + i| |z|^2 = |z + i|^2 Let z = a + bi |a + bi|^2 = |a + (b+1)i|^2 a^2 + b^2 = a^2 + (b+1)^2 a^2 + b^2 = a^2 + b^2 + 2b + 1 0 = 2b + 1 b = -1/2 So, z = a - i/2 Verification: |(a - i/2) / (a - i/2 + i)| |(a - i/2)/(a + i/2)| |(2a - i)/(2a + i)| |2a - i| / |2a + i| sqrt(4a^2 + (-1)^2) / sqrt(4a^2 + 1^2) sqrt(4a^2 + 1) / sqrt(4a^2 + 1) Which is 1, as 4a^2 + 1 can never be zero. So, this is true for all values of a.

[@Nobodyman181]

xy-xy=0

[@mcwulf25]

Yes, why didn't he cancel these and make that imaginary coefficient simpler? 9:51

[@ProficiencyMusic]

I did it in my head and got -i/2

[@kpt123456]

Congratulations

[@tolkienfan1972]

But there isn't only a single answer.

[@mcwulf25]

​@tolkienfan1972 you sure?

[@tolkienfan1972]

@@mcwulf25 did you watch the video?

[@lackadaisicalll]

a-i/2

[@sdspivey]

3rd method: Cross multiply, ±Z = Z+i Subtract Z from each side, so 0=i or -2Z=i First is not possible, so Z=-i/2 Just a single valid answer, not a range.

[@מיכאלקונטרוביץ]

This is totally wrong!!! You are taking the absolute value like a real number and you coincidentially get a part of a correct answer, but this is unacceptable

[@sdspivey]

@@מיכאלקונטרוביץ Then show me another value that works.

[@scottleung9587]

I also got z=-i/2, corresponding to the line y=-1/2.

[@JayTemple]

I got all numbers where the imaginary part is -1/2, but I didn't feel confident that there weren't other solutions.

[@tolkienfan1972]

z=-i/2 is a point, not a line

[@phill3986]

😊😊😊👍👍👍

[@sail2byzantium]

Well. I am glad somebody understood this . . . I lost the thread around 6:18 and why you square-rooted everything. And for such messy math, you should slow it down a bit. Things felt a bit rushed as to what exactly you were doing and why you were doing it.

[@aplusbi]

Thanks for the feedback