$|z_2|z_1 +|z_1|\bar z_2$

If for any two non zero complex numbers z1 and z2 , if $z_1 (1+|z_2|)=iz_2 (1+|z_1|)$ , then we have to find the value of $|z_2|z_1 +|z_1|\bar z_2$

My try

$z_1 (1+|z_2|)=iz_2 (1+|z_1|)$ in this I multiplies both sided by $\bar z_1$

from.that I got $\bar z_1 z_2$ is purely imaginary .

after that I got stuck

• How about $z_1=x_1+iy_1$ or $$z_1=r_1e^{it_1}$$ – lab bhattacharjee Apr 18 '17 at 16:10

Consider $z_1 = i$ and $z_2 = 1$.

Then, $z_1 (1+|z_2|) = i(1+1) = 2i$ and $iz_2(1+|z_1|) = i(1+1) = 2i$.

However, $|z_2|z_1 +|z_1|\overline{z_2} = |i|(i) + |i|\overline1 = i+1 \ne 0$.

I found exactly the same counterexample as DHMO. However, I want to share how I found it, so maybe you can come up with many more counterexamples.

I started setting $z_1=re^{i\theta},z_2=\rho e^{i\phi}$. The hypothesis then becomes:

$$re^{i\theta}(1+\rho)=i\rho e^{i\phi}(1+r).$$

Taking moduli, this yields:

$$r(1+\rho)=\rho(1+r)\iff r=\rho.$$

So $|z_1|=|z_2|$. Then I took arguments:

$$\theta=\phi+\frac{\pi}{2}+2k\pi\iff\theta-\phi=\frac{\pi}{2}+2k\pi.$$

Since this approach didn't seem to be leading me to a proof, I tried a disproof. Since the moduli are equal, I thought, why not set them to 1, simplifying the equations? The hypothesis, with moduli equal to 1, becomes:

$$2z_1=2iz_2\iff z_1=iz_2\iff z_1=e^{i\theta},z_2=e^{i(\theta-\frac{\pi}{2})}.$$

Plug into thesis:

$$0=z_1+\overline{z_2}=e^{i\theta}+e^{i(\frac{\pi}{2}-\theta)}=e^{i\theta}+ie^{-i\theta}=z_1+i\overline{z_1}.$$

Let's switch to algebraic form: $z_1=a+ib$. This gives, plugging into thesis:

$$0=a+ib+i(a-ib)=a+b+i(a+b)\iff a=-b.$$

So $z_1$ should be a real multiple of $1-i$, which with modulus 1 implies:

$$z_1=\frac{1}{\sqrt2}(1-i).$$

So just take any other $z_1$ with modulus one, and you will find a counterexample. For example, $z_1=i$. Plug into hypothesis to find $z_2$, always under the assumption that $|z_2|=|z_1|=1$:

$$0=z_1(1+|z_2|)-iz_2(1+|z_1|)=2i-2iz_2\iff z_2=1.$$

Plug into thesis, it doesn't work. Many more counterexamples can be found.

Note that, since the moduli are equal, you can just assume $|z_1|=r$ for any $r$, and the thesis and hypothesis merely get multiplied by $r^2$, so you still get $z_1$ is a multiple of $1-i$, which is very restrictive.

From $z_1 (1+|z_2|)=iz_2 (1+|z_1|)$ you have that $(1+|z_2|)\Im (z_1) = (1+|z_1|) \Re (z_2)$ and $(1+|z_2|)\Re (z_1) = - (1+|z_1|) \Im (z_2)$ .

Plugging in our two findings from above gives $$|z_2|z_1 +|z_1|\bar z_2=\\ |z_2|(- \frac{1+|z_1|}{1+|z_2|} \Im (z_2) + i \frac{1+|z_1|}{1+|z_2|} \Re (z_2)) +|z_1|(\Re (z_2) - i \Im (z_2))=\\ (- |z_2| \frac{1+|z_1|}{1+|z_2|} \Im (z_2) + |z_1|(\Re (z_2) ) - i (- |z_2|\frac{1+|z_1|}{1+|z_2|} \Re (z_2)) + |z_1| \Im (z_2))$$

Further, from $z_1 (1+|z_2|)=iz_2 (1+|z_1|)$ you have that $|z_1| (1+|z_2|)=|z_2| (1+|z_1|)$ from which you even have $|z_1| =|z_2|$. So our expression reduces to

$$|z_2|z_1 +|z_1|\bar z_2=\\ |z_1|\Big[ ( \Re (z_2) - \Im (z_2) ) + i ( \Re (z_2) - \Im (z_2)) \Big] =\\ |z_2|\Big[\bar z_2 + i z_2 \Big]$$

Hint: taking the square of magnitudes on both sides of $\,z_1 (1+|z_2|)=iz_2 (1+|z_1|)\,$:

$$\require{cancel} |z_1|^2(1+|z_2|)^2=|z_2|^2(1+|z_1|)^2 \\ |z_1|^2(1 + 2 |z_2| + \bcancel{|z_2|^2})=|z_2|^2(1+2|z_1|+\bcancel{|z_1|^2}) \\ (|z_1|-|z_2|)(|z_1|+|z_2|+2 |z_1||z_2|)=0$$

The latter gives $|z_1|=|z_2|\,$, so the given relation simplifies to $z_1=i z_2\,$.

Then $|z_2|z_1 +|z_1|\bar z_2 = |z_2|(i z_2 + \bar z_2)=|z_2|\cdot \left(\operatorname{Re}(z_2)-\operatorname{Im}(z_2)\right)\cdot(1+i)\,$.