Complex numbers equation with modulus and De Moivre's formula

Question

I have a problem with the following question.

For which $n$ does the following equation have solutions in complex numbers

$$|z-(1+i)^n|=z $$

Progress so far.

1. Let $z=a+bi$.

2. Since modulus represents a distance, the imaginary part of RHS has to be 0. This immediately makes $b=0$.

3. If solutions are in the complex domain $|a-(1+i)^n|=a $ by 2., and $a$ is Real.

4. ?

I don't know where to go from here.

Answer 1

Since $z$ is equal to the absolute value. Therefore $z$ has to be real. Thus it has no purely imaginary solutions.

Answer 2

Note that $1+i=\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4))$. We have

$|z-(1+i)^n|=|z-(\sqrt{2})^n(\cos(n\pi/4)+i\sin(n\pi/4))|=z$

$\sqrt{(z-(\sqrt{2})^n\cos(n\pi/4))^2+(\sqrt{2})^{2n}\sin^2(n\pi/4)}=z$

so

$(z-2^{n/2}\cos(n\pi/4))^2+2^n\sin^2(n\pi/4)=z^2$,

$-2^{n/2+1}z\cos(n\pi/4)+2^n\cos^2(n\pi/4)+2^n\sin^2(n\pi/4)=0$,

$2^{n/2+1}z\cos(n\pi/4)=2^n$,

$$z=\frac{2^{n/2-1}}{\cos(n\pi/4)}$$

whenever $n$ is odd; for $n$ even there is no solution.

Answer 3

$|z-(1+i)^n|=z$. LHS $\in \mathbb{R}$ so clearly $z \in \mathbb{R}$. I'll rewrite $z$ as $a$. $(1+i)^n =(\sqrt2)^ne^{\frac{in\pi}{4}}$.

Thus, RHS is the square root of $(a-(\sqrt2)^n\cos(\frac{n\pi}{4}))^2+((\sqrt2)^n\sin(\frac{n\pi}{4}))^2$ (it is getting cumbersome so I didn't include the square root).

Thus we get $a^2 -2a(\sqrt2)^n\cos(\frac{n\pi}{4})+2^n\cos^2(\frac{n\pi}{4})+2^n\sin^2(\frac{n\pi}{4})=a^2$. After some algebraic manipulation:

$a(\sqrt2)^n\cos(\frac{n\pi}{4})=2^{n-1}$ and thus $a=\frac{2^{n/2-1}}{\cos(\frac{n\pi}{4})}$.

By the way, $\cos(\frac{n\pi}{4})=1,\frac{1}{\sqrt2},0,\frac{-1}{\sqrt2},-1,\frac{-1}{\sqrt2},0,\frac{1}{\sqrt2}$ for $n=0,...,7$ before repeating.

So given an $a$, you can find the zeros of $a-\frac{2^{n/2-1}}{\cos(\frac{n\pi}{4})}$. But based on what we know about $\cos(x)$, not all values of $a$ will work and also some values of $n$ will make the denominator equal to $0$.

Answer 4

You already found that z has to be real, now just write out $|z| = \sqrt{Re(z)^2 + Im(z)^2}$
Note that $i+1 = \sqrt{2} e^{i\frac{\pi}{4}} = \sqrt{2} \cos(\frac{\pi}{4}) + i\sqrt{2} \sin(\frac{\pi}{4})$
Then you need to find when $a^2 =|a-(1+i)^n|^2 = a^2 + 2 \cdot 2^\frac{n}{2} a \cos(\frac{\pi}{4}n) + 2^n$
That is, finding the zeroes of $2 \cdot 2^\frac{n}{2} a \cos(\frac{\pi}{4}n) + 2^n$
For small z this has solutions around $n \approx -2-4k$ (because the cos is zero and $2^n$ is negligible) but the only way this can be true for all z is in the limit $n\rightarrow - \infty$

Answer 5

Let's break the problem into two parts. (I) Find the set $S$ of all complex numbers $z_0$ such that the equation $|z-z_0|=z$ has a solution. (II) Determine the set of all integers $n$ such that $(1+i)^n\in S$.

(I) The equation $|z-z_0|=z$ says that $z$ is a nonnegative real number and is equidistant from $z_0$ and $0$. In the trivial case $z_0=0$, every nonnegative real $z$ satisfies the equation. Otherwise, the condition for the equation to be solvable is that the perpendicular bisector of the straight line segment joining $z_0$ to $0$ intersects the positive real axis. Clearly, this holds if and only if $z_0$ lies to the right of the imaginary axis. Thus $S=\{0\}\cup\{z:\Re(z)\gt0\}$.

(II) Since $(1+i)^n=2^{n/2}(\cos\frac{n\pi}4+i\sin\frac{n\pi}4)$, we have $(1+i)^n\in S$ just when the angle $\frac{n\pi}4$ is in the open right half-plane, that is, when $n=8t$ or $n=8t\pm1$, $t=0,\pm1,\pm2,\pm3,\dots$.

If $n=8t$, then $(1+i)^n=2^{4t}$, and the solution of the equation is $z=\frac12(1+i)^n=2^{4t-1}$, the midpoint of $0$ and $(1+i)^n$.

If $n=8t\pm1$, then $(1+i)^n=2^{4t\pm\frac12}(\cos\frac\pi4\pm i\sin\frac\pi4)$, and the solution of the equation is $z=\Re(1+i)^n=2^{4t\pm\frac12}\cos\frac\pi4=2^{4t-\frac12\pm\frac12}$, i.e., $z=2^{4t}$ if $n=8t+1$ or $z=2^{4t-1}$ if $n=8t-1$, in either case making an isosceles right triangle with $0$ and $(1+i)^n$.

Answer 6

• Obviously, z is a point on the positive semi-axis, since absolute values are always positive reals.
• $|z_1-z_2|$ represents the distance between two points on the complex plane.
• Therefore, z is equally distanced from the origin O, and from $A_n=(1+i)^n$. This implies that it lies on the perpendicular bisector of the segment $(OA_n)$.
• Since n is a natural number, and since the angle between $A_1=1+i$ and the positive semiaxis is $45^\circ$, it follows that, with each successive multiplication, the angle between the segment $OA_k$ and the positive semiaxis grows by $45^\circ$, until it resets on the eighth turn, since $360^\circ/45^\circ=8$.
• But on which of these eight cyclically repeating situations does the perpendicular bisector of the segment $(OA_n)$ actually meet or intersect the positive semiaxis ? Obviously, it can only happen for $n\equiv0,\pm1\mod8$.