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$\arg(z_1+z_2)=0$ and $\text{Im}(z_1z_2)=0$. Prove that $z_1=\bar{z_2}$

I substituted $z_1=a+ib$ and $z_2=c+id$

Using first condition, $$z_1+z_2=\bar{z_1}+\bar{z_2}$$ $$a+ib+c+id=a-ib+c-id$$ $$b+d=0$$

So $z_1=a+ib$, $z_2=c-ib$

Using second condition, $$\text{Im}(ac-iab+ibc+b^2)=0$$ If $b\neq 0$ $$a=c$$

So $z_1=a+ib$ and $z_2=a-ib$

Is there a way without substitution?

Aditya Dev
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2 Answers2

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Saying that $\arg(z_1+z_2)=0$ means that $z_1+z_2$ is real and nonnegative. In particular the imaginary parts sum to $0$ and for the real parts you have $a+c\ge0$.

The second condition says $ad+bc=0$, which together with $d=-b$ tells you $b(a-c)=0$. Thus either $b=0$ (so also $d=0$), or $a=c$.

Thus, the statement is false, because obviously $z_1=1$ and $z_2=2$ satisfy the requirements, but are not conjugate of each other.

Therefore, no: there no way to prove the result, with or without substitutions.

If one assumes $z_1$ and $z_2$ nonreal, the result is true. The first condition implies $z_1+z_2=\bar{z}_1+\bar{z}_2$, the second condition that $z_1z_2=\bar{z}_1\bar{z}_2$. Therefore $$ 0=z_1z_2-\bar{z}_1\bar{z}_2= z_1z_2-z_1\bar{z_2}+z_1\bar{z_2}-\bar{z}_1\bar{z}_2= z_1(z_2-\bar{z}_2)+(z_1-\bar{z}_1)\bar{z}_2 $$ Applying the fact that $z_2-\bar{z}_2=-(z_1-\bar{z}_1)$, we get $$ (z_1-\bar{z}_1)(z_1-\bar{z}_2)=0 $$ Since $z_1-\bar{z}_1\ne0$, we deduce $z_1-\bar{z}_2=0$.

egreg
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Since $arg(z_1+z_2)=0$ you have $$z_1+z_2=\bar{z_1}+\bar{z_2}=:a$$

Since $Im(z_1z_2)=0$ you have $$z_1z_2=\bar{z_1} \bar{z_2}=:b$$

Now, consider the quadratic equation $$z^2-az+b=0$$

Then $z_1,z_2$ are the two roots. Also $\bar{z_1}, \bar{z_2}$ are the two roots.

This proves that $$z_1=\bar{z_1} \,;\, z_2 =\bar{z_2} \mbox{ or } \\ z_1=\bar{z_2} \,;\, z_2 =\bar{z_1} $$

The first case provides a counterexample, but under the extra assumption that $z_1$ is not real, you get the desired conclusion.

N. S.
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