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Let z be the complex number of maximum amplitude (argument) satisfying $$|z-3|=Re(z),$$ then I need the value of $|z-3|$

So I proceeded with substituting $z=x+iy$, and got the following:-

$$\sqrt{(x-3)^2+y^2}=x$$, and squaring it I got the equation of a parabola. Then how do I proceed? I know that I need that complex number $z$ that has the maximum amplitude, that is, the one that makes the maximum positive angle with the $x$-axis.
How do I do that and then subsequently, how do I find the value of $\vert z-3\vert$?

TheHolyJoker
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Techie5879
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  • To explain the appearance of the parabola, note that your equation specifies that the distance of $z$ from the imaginary axis (ie the real part of $z$) be the same as the distance between $z$ and $3$. But that’s precisely the directrix-focus definition of a parabola. – Semiclassical Dec 20 '19 at 18:38
  • There is no solution. The point $({1 \over 6}(t^2-9),t)$ satisfies the requirement and has arbitrary length. – copper.hat Dec 20 '19 at 18:43
  • Oops, $+9$ not $-9$. – copper.hat Dec 20 '19 at 18:54

2 Answers2

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Hint

$$y^2=6x-9$$

We need to maximize $$\dfrac yx=\dfrac{6y}{y^2+9}=1-\dfrac{(y-3)^2}{y^2+9}\le1$$

Do you when does the equality occur?

Alternatively for $ay>0$ , $$\dfrac{ay}{a^2+y^2}=\dfrac1{\dfrac ay+\dfrac ya}$$

and using AM-GM inequality, $$\dfrac{\dfrac ay+\dfrac ya}2\ge\sqrt{\dfrac ay\cdot\dfrac ya}=1$$

lab bhattacharjee
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The focus-directrix property can give you the result without even finding the equation.

The required point is directly above the focus i.e. at $(3,3)$. This point is where the tangent to the ellipse has gradient $1$ and passes through the origin.

Therefore $|z-3|=3$.