alinazarboland wrote: find all $f : \mathbb{C} \to \mathbb{C}$ st: $$f(f(x)+yf(y))=x+|y|^2$$for all $x,y \in \mathbb{C}$ 1) Lemma : Find all $g(x)$ from $\mathbb R\to\mathbb C$ such that $Q(x,y)$ : $g(x+y^2)=g(x)+yg(y)$ $\forall x,y\in\mathbb R$

2) Original problem 2.1) $f(x)=cx$ $\forall x\in\mathbb R$

2.2) $f(x)$ over $\mathbb C$

And so $\boxed{f(x)=e^{i\theta}\overline x\quad\forall x\in\mathbb C}$, which indeed fits, whatever is $\theta\in[0,2\pi)$

Let $P(x,y)$ be the assertion $f(f(x)+yf(y))=x+|y|^2$. $P(x,0) \implies f(f(x))=x$. With this $P(x,f(y)) \implies |y| = |f(y)|$ for all $y\in \mathbb{C}$ which leads us to $f(0)=0$. Let $|y_{1}|=|y_{2}|$ be two complex numbers. Then using injectivity of $f$ we get $P(x,y_{1}), P(x,y_{2}) \implies y_{1}f(y_{1}) = y_{2}f(y_{2})$. Let $y_{1} =y$ and $y_{2} = |y|$ we get that $$f(y) = \dfrac{\bar{y}}{|y|}f(|y|)\; \;\;\;(1).$$ So it suffices to find $f$ on the real line $\mathbb{R}$. By applying $f()$ both sides of $P(x,y)$ we get $f(x+|y|^2)=f(x)+yf(y)$. By inserting $x=0$ in this equation and rewriting the equation we get $f(x+|y|^{2})=f(x)+f(|y|^{2})$. Then for any two positive real numbers $x,y$ we have $$f(x+y)=f(x)+f(y) .$$We know that $|f(x)|=|x|$ for any complex number $x$. Using triangle inequality, for positive reals $a,b$ we have $$a+b = |a+b| = |f(a+b)| = |f(a)+f(b)| \leq |f(a)| + |f(b)| = |a| + |b| = a+b.$$The equality case of triangle inequality gives us that $f(x)=xf(1)$ for all positive real $x$. So $f(|x|)= |x|f(1)$ for any $x\in \mathbb{C}$. This with $(1)$ gives us $$\forall y\in \mathbb{C}:\;f(y) = f(1)\cdot \bar{y}.$$And so $\boxed{f(y)=e^{i\theta}\overline y\quad\forall y\in\mathbb C}$, which indeed fits, whatever is $\theta\in[0,2\pi)$.

$P(x,y):f(f(x)+yf(y))=x+|y|^{2}$ so $P(x,0):f(f(x))=x$ As a result by setting $y=f(y)$ yields: $$x+|y|^{2}=f(f(x)+yf(y))=f(f(x)+f(f(y))f(y))=x+|f(y)|^{2}$$So $|y|=|f(y)|$ setting y=0 yields $|f(0)|=0$ so $f(0)=0$ Now we know $P(0,y):f(yf(y))=|y|^{2}$ so $yf(y)=f(f(yf(y)))=f(|y|^{2})$ so $f(y)=\frac{f(|y|^{2})}{y}$ As a result for all positive real number $x$ we know :$$P(x,1):\frac {f(|x+1|^2)}{f(x)+f(1)}=\frac{f(|f(x)+f(1)|^2)}{f(x)+f(1)}=f(f(x)+f(1))=x+1$$So $f(x^{2} +2x+1)=xf(x)+xf(1)+f(x)+f(1)$ (we call it result 1) Now by using the original equation we have $$f(x)+f(|y|^2)=f(f(f(x)+f(|y|^2)))=f(x+|y|^2) $$so for any positive real numbers x ,y we understand $f(x+y)=f(x)+f(y)$ Now with this result and result 1 for any positive real number x we know $f(x)=xf(1)$ so for any complex number x we have $f(x) = f(1)|x|^2 /x = f(1)\overline{x}$ by checking the answer in the original equation we understand only possible answers are $f(x)=b\overline{x}$ how b is a complex number and $|b|=1$

Solved with megarnie. First, taking $y = 0$ gives that $f$ is an involution, and hence injective. Taking $x = 0, y = f(0)$ then gives $|f(0)|^2 = 0$ $\implies f(0) = 0$. Then taking $x = 0$ again we get $$f(yf(y)) = |y|^2\qquad (\spadesuit)$$Replacing $y$ by $f(y)$ in $(\spadesuit)$ gives $f(f(y)y) = |f(y)|^2$. Comparing this and $(\spadesuit)$ gives $|f(y)| = |y|$. Applying $f$ to both sides of $(\spadesuit)$, we also get that $f(|y|^2) = yf(y)$. Hence, the original FE becomes $f(f(x) + f(|y|^2)) = x + |y|^2$. Now take $x = 1$ and we get $f(f(1) + f(|y|^2)) = 1 + |y|^2$. Taking the magnitude of both sides, we have $|f(1) + f(|y|^2)| = 1 + |y|^2$. But from the triangle inequality we have \[ |f(1) + f(|y|^2)|\le |f(1)| + |f(|y|^2)| = 1 + |y|^2,\]so we must have the equality case, which is when $f(1)$ and $f(|y|^2)$ have the same argument. This means $f(r)$ for reals $r > 0$ all have the same argument, say $k$. We also know $|f(r)| = |r| = r$ for reals $r > 0$, so we actually must have $f(r) = e^{ik}r$ for reals $r > 0$. Now, consider the equation $f(|y|^2) = yf(y)$ again. With our new information, the LHS is $e^{ik}|y|^2$. Hence, $\boxed{f(y) = e^{ik}|y|^2/y\quad \forall y\ne 0,\text{ and }f(0) = 0}$, or equivalently $\boxed{f(y)\equiv e^{ik}\bar{y}}$. It's easy to check that this works by plugging it into the FE.

We claim the only functions are $f(x)\equiv c\overline{x}$ where $|c|=1$. It is easy to check that these work. Let $P(x,y)$ denote the given assertion. From $P(x,0)$ we get $f(f(x))=x$ so $f(f(0))=0$. From $P(f(0),f(0))$ we get $f(0)=f(0)+|f(0)|^2$ so $f(0)=0$. We have \begin{align*} P(0,x)&\implies f(xf(x))=|x|^2\tag{1}\\ P(0,f(x))&\implies f(xf(x))=|f(x)|^2 \end{align*}so $|f(x)|=|x|$. Applying $f$ to both sides of $(1)$ gives \[ xf(x)=f(|x|^2).\tag{2} \]Then $P(x,y)$ becomes $f(f(x)+f(|y|^2))=x+|y|^2$ and applying $f$ to both sides gives $f(x+|y|^2)=f(x)+f(|y|^2)$. Thus \[ f(x+y)=f(x)+f(y)\ \ \ \ \ \ \forall (x,y)\in\mathbb{C}\times\mathbb{R}_{\geq 0}.\tag{3} \]Taking $x\in\mathbb{R}_{<0}$ and $y=-x\in\mathbb{R}_{\geq 0}$ gives $f(x)=-f(-x)$ for all real $x$. Claim. For all real $x$ and $y$, we have \[ f(x+y)=f(x)+f(y). \] Proof. If at least one of $x$ or $y$ is nonnegative, WLOG suppose $y\geq 0$. Then we are done by $(3)$. If $x,y<0$, we have \[ f(x+y)=-f(-x-y)=-f(-x)-f(-y)=f(x)+f(y) \]and we are done. $\square$ Let $g:\mathbb{R}\rightarrow\mathbb{R}$ and $h:\mathbb{R}\rightarrow\mathbb{R}$ be given by $g(x):=\mathrm{Re}(f(x))$ and $h(x):=\mathrm{Im}(f(x))$. Note that $g(x+y)=g(x)+g(y)$ and $h(x+y)=h(x)+h(y)$ for all real $x$ and $y$. Since $g(x)^2+h(x)^2=|f(x)|^2=|x|^2$, it follows that $|g(x)|,|h(x)|\leq |x|$ for all $x$. Since $g$ and $h$ are bounded in $[665,1434]$, they must be linear. Thus $g(x)\equiv c_1x$ and $h(x)\equiv c_2x$ for some real $c_1$ and $c_2$ so $f(x)=(c_1+c_2i)x=cx$ for all real $x$. By $(2)$, we get that $f(x)=\frac{f(|x|^2)}{x}=\frac{c|x|^2}{x}=c\overline{x}$ for all $x\in\mathbb{C}\setminus\{0\}$. Since $|f(x)|=|x|$, it follows that $|c|=1$. $\square$

The only functions that work are $f(x)=z\overline{x}$ for some complex number $z$ with $|z|=1$. To verify that these functions work notice that $$f(f(x)+yf(y))=f(z\overline{x}+z|y|^2)=x+|y|^2$$Now let $P(x,y)$ be the given assertion. Claim: $f(f(x))=x$ Follows from $P(x,0)$. Claim: $|y|=|f(y)|$ This follows from comparing the assertions for $P(x,y)$ and $P(x,f(y))$. Notice that this implies that $f(0)=0$. Claim: $yf(y)=f(|y|^2)$ The assertion $P(0,y)$ gives that $f(yf(y))=|y|^2$ so $f(f(yf(y)))=yf(y)=f(|y|^2)$. Finishing: Let $f(1)=z$. Now let $\mathcal{D}\subseteq \mathbb{C}$ denote the set of all complex numbers $x$ such that $f(x)=z\overline{x}$. Notice that $1\in \mathcal{D}$ and by the above claim, if $y\in\mathcal{ D}$ then all complex numbers $x$ with $|x|=|y|$ are in $\mathcal{D}$. Furthermore, if $x,y \in\mathcal{ D}$ then $f(x)+yf(y)=z\overline{x}+z|y|^2$ is also in $\mathcal{D}$. Now notice that we have that all complex numbers $x$ with $|x|=1$ are in $\mathcal{D}$. Also if we choose $|x|=|y|=1$ then $z\overline{x}$ can take on any unit complex number and thus $a+z\in \mathcal{D}$ for all unit complex numbers $a$. This achieves all magnitudes in $[0,2]$ thus the closed disk with radius $2$ centered about the origin belongs in $\mathcal{D}$. By a simple induction argument we can increase the radius of this disk by one and still have it lie in $\mathcal{D}$, thus $\mathbb{C}=\mathcal{D}$ finishing the problem.

Cute. Would give headpats. The only functions that work are $f(x) = \omega \bar x$ for $|\omega| = 1$. $y=0$ gives us that $f(f(x)) = x$. Notably, swapping $y$ and $f(y)$ we obtain that $|f(y)| = |y|$. Let $\omega = f(1)$ and suppose $\omega \neq \pm 1$. Note that $|\omega| = 1$. Then \[ f(f(x) + \omega) = x + 1. \]We then require some complex $z := f(x)$ with $|z| = |x|$ such that $|z + \omega| = |x + 1|$, or $|\frac{z}{\omega} + 1| = |x + 1|$ which implies that $f(x) = \omega x$ or $f(x) = \omega \bar{x}$ for each $x$. This follows from geometry. If $f(x) = \omega x$, then $x = f(\omega x)$, which is either $\omega^2 x$ or $\omega \bar{\omega x} = \bar x$, neither of which are possible given our assumption on $\omega$. Hence if $\omega \neq -1$ we have our function. Now suppose $\omega = -1$. Then \[ f(yf(y)) = |y|^2. \]Now suppose that $y$ is not real, and that $f(y) = -y$. Then $f(-y^2) = |y|^2$, impossible unless $y^2$ is real, i.e. $y$ is real or $y$ is purely imaginary. But if $y$ is purely imaginary, and $f(y) = -y$, then \[ f(y - 1) = f(y) + 1 = -y+1, \]a contradiction. Suppose $\omega = 1$ and say for nonreal $y$ we have $f(y) = y$. Then \[ f(f(x) + y^2) = x + |y|^2. \]If $x$ is and $y$ isn't purely real or purely imaginary, the arguments don't match up, bad. Now if $y$ is purely imaginary, again $f(y + 1) = f(y) + 1 = y + 1$, which we plug back in to obtain a contradiction.