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If $z_1,z_2$ are two complex numbers such that $\vert z_1+z_2\vert=\vert z_1\vert+\vert z_2\vert$,then it is necessary that

$1)$$z_1=z_2$

$2)$$z_2=0$

$3)$$z_1=\lambda z_2$for some real number $\lambda.$

$4)$$z_1z_2=0$ or $z_1=\lambda z_2$ for some real number $\lambda.$

From Booloean logic we know that if $p\implies q$ then $q$ is necessary for $p$.

For $1)$taking $z_1=1$ and $z_2=2$ then $\vert 1+2 \vert=\vert 1\vert+\vert 2\vert$ but $1\neq 2$.So,$(1)$ is false.

For $2)$taking $z_1=1$ and $z_2=2$ then $\vert 1+2 \vert=\vert 1\vert+\vert 2\vert$ but $2\neq 0$.So,$(2)$ is false.

I'm not getting how to prove or disprove options $(3)$ and $(4)?$

Need help

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    $\begingroup$ $z_1=1,z_2=0$ shows that 3) is false. $\endgroup$ – Kavi Rama Murthy Jun 4 '18 at 7:27
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A bit of geometry:

Consider $z_1, z_2$ vectors in the complex plane.

Vector addition:

$\vec {OA} =z_1$; $\vec {AB} =z_2$; and $\vec {OB} = z_1+z_2.$

For $OAB$ to be a triangle we must have:

$|z_1|+|z_2|> |z_1+z_2|$, i.e. the sum of the lengths of 2 sides is greater than the 3rd side.

Hence $z_1$ and $z_2$ must be collinear.

1) $\lambda =0$, trivial.

2)$\lambda <0:$

$z_1=\lambda z_2.$

$|1+\lambda| |z_1| = (1+|\lambda|)|z_1|.$

Ruled out.

3) $\lambda >0$ , ok.

Hence a necessary condition :

$z_1= \lambda z _2$ with $\lambda \ge 0$.

Now check your options.

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Hint:

3) Consider $\lambda=-1$ for some $z_1\neq 0$.

4) Complex numbers forms a field. So $z_1z_2=0$ implies $z_1=\ldots$ or $z_2=\ldots$. Hence, 4) is the same as $\ldots$ or 3)

Nevertheless, 3) would be true, if $\lambda$ were a real and positive number.

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$$\vert z_1+z_2\vert=\vert z_1\vert+\vert z_2\vert \Rightarrow \vert z_1+z_2\vert^2= (\vert z_1\vert+\vert z_2\vert)^2 \Rightarrow Re(z_1\bar z_2) = |z_1||z_2|$$ Now, note that $Re(z_1\bar z_2)$ is a real 2-dimensional scalar product $$Re(z_1\bar z_2) = x_1x_2+y_1y_2 \mbox{ where } z_1 = x_1+iy_1, \, z_2 = x_2+iy_2$$ Cauchy-Schwarz tells us that $$Re(z_1\bar z_2) = |z_1||z_2| \Rightarrow z_1z_2= 0 \mbox{ or } z_1 = \lambda z_2 \mbox{ where } \lambda > 0$$

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In general, the numbers $z_1,\,z_2,\,z_1+z_2$ represent vectors $\vec{u},\,\vec{v},\,\vec{u}+\vec{v}$ starting from $O.$ Thus is $|z_1+z_2|$ the length of a diagonal in the parallelogram with the sides $|z_1|$ and $|z_2|.$ Due the triangular inequality is $$|z_1+z_2|\leq |z_1|+|z_2|.$$ The equality occurs only in the degenerate case(s), where the vertices are collinear. This gives the necessary condition:

All four vertices of the parallelogram are equal OR they are different but collinear.

The answer 4) is right.

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