Show complex solutions exist

Question

Let A be a complex number and B a real number. Show that the equation $\,\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B = 0\,$ has a solution iff $\,\lvert A^2\rvert \geq 4B$. If this is so, show that the solution set is a circle or a single point.

Well i am trying to do the first part first. So assuming the equation has a solution that would mean $z = x+iy$ satisfies the equation.

I was going to let $A = s+it$ for a complex number, but it is not working out for me. Wrong step?

You surely meant "a real solution", as any polynomial equation has a solution in the complex field. — Timbuc, Oct 26 '14 at 17:32
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.10? — BCLC, Jul 29 '18 at 12:26

score 2 · Accepted Answer · answered Oct 26 '14 at 17:32

2

We have $$ 0=\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B = z\overline{z}+\frac{1}{2}(Az+\overline{Az})+\frac{1}{4}\lvert A\rvert^2-\frac{1}{4}\lvert A\rvert^2+B=\left\lvert z+\frac{1}{2}A\right\rvert^2-\frac{1}{4}\lvert A\rvert^2+B $$ If $4B>\lvert A\rvert^2>0$, then $\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B=\left\lvert z+\frac{1}{2}A\right\rvert^2-\frac{1}{4}\lvert A\rvert^2+B>0$, and hence no solutions.

If $4B>\lvert A\rvert^2\le 0$, set $C=\frac{1}{2}\sqrt{\lvert A\rvert^2-4B}$, and our equation is equivalent to $$ \left\lvert z+\frac{1}{2}A\right\rvert^2=C^2, $$ and hence equivalent to $$ \left\lvert z+\frac{1}{2}A\right\rvert=C, $$ the set of solutions of which is the circle centered at $-A/2$ with radius $C$.

answered Oct 26 '14 at 17:32

Yiorgos S. Smyrlis

83,933

Everything makes sense in all of these solutions except for one minor detail. |z^2| = |z|^2 ? – D.C. the III Oct 26 '14 at 17:52
Which minor detail? – Yiorgos S. Smyrlis Oct 26 '14 at 17:53
I thought you could not carry the exponential through like that with absolute values. – D.C. the III Oct 26 '14 at 17:55
@dc3rd let $z = x+iy$ and work out $|z^2|$ and $|z|^2$. you can prove it to yourself. – dustin Oct 26 '14 at 17:58
Well unless i am missing a last step in the simplification of |z^2| then i am not getting the same answer – D.C. the III Oct 26 '14 at 18:12
$\lvert z^2\rvert=\lvert x^2-y^2+2xyi\rvert=\sqrt{(x^2-y^2)^2+(2xy)^2}=\sqrt{(x^2+y^2)^2}=x^2+y^2=(\sqrt{x^2+y^2})^2=\lvert z\rvert^2$. – Yiorgos S. Smyrlis Oct 26 '14 at 18:19
I made a mistake. I wrote x^2 + y^2. Tganks for clearing that up. Guys like you are what make this place tick. I just hope i can soon be in a position to do the same for others. Im going to work on the rest of this and hopefully it wotks out. Thanks – D.C. the III Oct 26 '14 at 18:30

score 1 · Answer 2 · answered Oct 26 '14 at 17:22

1

You are on the right track by letting $z = x+iy$ and $A=s+it$. If substitute this in, we have $$ x^2+y^2+xs - yt+r = 0 $$ where $r$ is my real number. Try completing the square for $x$ and $y$ and see what you get.

answered Oct 26 '14 at 17:22

dustin

8,241

score 0 · Answer 3 · answered Oct 26 '14 at 17:23

Let $z=x+iy, A=a+ib$ where $x,y,a,b$ are real

So, we have $x^2+y^2+ax-by+B=0$

$\implies\left(x+\dfrac a2\right)^2+\left(y-\dfrac b2\right)^2=B-\dfrac{a^2+b^2}4$

But $\left(x+\dfrac a2\right)^2+\left(y-\dfrac b2\right)^2\ge0$

Hope you can take it from here

Show complex solutions exist

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