Question

|Z1| = |Z2| = 1 then prove that |Z1 +Z2| = |1\Z1 + 1\ Z2|​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: If \: z_1 \: and \: z_2 \: are \: complex \: numbers \: such \: that \: |z_2| = |z_1| = 1$

$\sf \: then \: prove \: that \: |z_1 + z_2| = |\dfrac{1}{z_1} + \dfrac{1}{z_2} |$

$\large\underline{\bold{Solution-}}$

Identity used :-

$\bull \: \sf \: If \: z \: = a + ib, \: then \: conjugate \: \: \overline{z} = a - ib$

$\bull \sf \: If \: z \: is \: complex \: number, \: then \: z \overline{z} = { |z| }^{2}$

$\bull \: \sf \: If \: z \: is \: complex \: number, \: then \: |z| = |\overline{z}|$

Let's solve the problem now!!

Given that

$\sf \: |z_1| = 1$

On squaring both sides, we get

$\sf \: { |z_1| }^{2} = 1$

$\sf \: z_1\overline{z_1} = 1 \: \: \: \: \: \: \: \: \: \: \: ( \bf\because \: { |z| }^{2} = z\overline{z})$

$\sf \: \therefore \: z_1 = \dfrac{1}{\overline{z_1}} - - (1)$

Similarly,

$\sf \: \therefore \: z_2 = \dfrac{1}{\overline{z_2}} - - (2)$

Now,

• Consider,

$\sf \: |z_1 + z_2|$

$\sf \: = \: \bigg|\dfrac{1}{\overline{z_1}} + \dfrac{1}{\overline{z_2}} \bigg| \: \: \: \: \: \: \{ \: \bf \: using \: equation \: (1) \: and \: (2) \}$

$\sf \: = \: \bigg|\overline{\dfrac{1}{z_1} + \dfrac{1}{z_2} } \bigg| \: \: \: \: \: \: \{ \bf\because \: \overline{z_1 + z_2} = \overline{z_1} + \overline{z_2} \}$

$\sf \: = \: \bigg|\dfrac{1}{z_1} + \dfrac{1}{z_2} \bigg| \: \: \: \: \: \: \: \{ \bf\because \: |\overline{z}| = |z| \}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

Conjugate of Complex Number :-

$\sf \: Conjugate \: of  \: a \: complex \: number \:  z = x + iy  \:  is   \: x − iy$

$\sf \: and \: which \: is \: denoted \: as \: \overline{z}$

That is,

$\sf \: If \: z = \: x + iy \: then \: = x - iy$

Properties of conjugate :-

$\sf \: \overline{z_1 - z_2} = \overline{z_1} - \overline{z_2}$

$\sf \: \overline{z_1 z_2} = \overline{z_1} \: \: \overline{z_2}$