Question

\[z, A\in \mathbb C, b\in \mathbb R\]
\[|z|^2+Re(Az)+b=0\text{ has a solution }\iff |A|^2\geq 4b\]

Answer 1

This is a slightly obscure question. Is this a result you need as a lemma of a larger proof?

Answer 2

this is the entire question

Answer 3

and just to be clear, even though A is in caps, it is also just a complex number, yes?

Answer 4

\[z=a+bi,A=e+fi\]
\[a^2+b^2+ae-bf+b=0\]
yes

Answer 5

A in C say
\[A=e+fi\]

Answer 6

\[a^2+b^2\geq0\]
\[ae-bf+d\leq0\]
\[d\leq ae-bf\] stuck

Answer 7

sorry i made a mistake in the question.

Answer 8

Note that your last equation should be bf - ae \geq b.

Answer 9

\[|z|^2+Re(Az)+d=0\]

Answer 10

Ok and d is real.

Answer 11

d real yes

Answer 12

and i assume the inequality involves d, not b.

Answer 13

yes sorry

Answer 14

\[|A|^2\geq 4d\]

Answer 15

I'll think about it during Doctor Who. :-)

Answer 16

ty

Answer 17

Ok, got it.

Answer 18

@jamesj i sure don't
love to see your answer...

Answer 19

Write z = R.exp(beta.i) and A = S.exp(alpha.i)

Then |z|^2 = R^2 and Re(zA) = RS.cos(alpha + beta)

Thus |z|^2 + Re(zA) + d = 0 is

R^2 + RS.cos(alpha + beta) + d = 0

and this quadratic has a solution if and only if the discriminent

S^2 cos^2(alpha + beta) - 4d >= 0

i.e.,

S^2 >= 4d, as alpha and beta are arbitrary

i.e.,

|A|^2 >= 4d

Answer 20

thank you so much