0
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ How about $z_1=x_1+iy_1$ or $$z_1=r_1e^{it_1}$$ $\endgroup$ – lab bhattacharjee Apr 18 '17 at 16:10
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.