Question

If x, y, z are three complex numbers such that
$|x| = 2 \: \: and \: \: |y| = 3 \: \: and \: \: |z| = 4 \: and \: |2x + 3y + 4z| = 4$

Then find the value of |8yz + 27zx + 64xy|

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Answer 1

Answer:

∣z

1

∣=9, ∣z

2

−(3+4i)∣=4

C

1

(0,0) radius r

1

=9

C

2

(3,4) radius r

2

=4

C

1

C

2

=∣r

1

−r

2

∣=5

∴ circle touches internally

∴ ∣z

1

−z

2

∣

min

=0

hope it helps you

Answer 2

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

Given that, x, y, z are complex numbers such that

$\rm :\longmapsto\: |x| = 2$

$\rm :\longmapsto\: |y| = 3$

$\rm :\longmapsto\: |z| = 4$

$\rm :\longmapsto\: |2x + 3y + 4z| = 4$

Now, Consider,

$\rm :\longmapsto\: |8yz + 27zx + 64xy|$

can be rewritten as

$\rm \:  =  \: \bigg |\dfrac{8xyz}{x} + \dfrac{27xyz}{y} + \dfrac{64xyz}{z} \bigg |$

can be further rewritten as by taking xyz common

$\rm \:  =  \: \bigg |xyz\bigg(\dfrac{8}{x} + \dfrac{27}{y} + \dfrac{64}{z}\bigg) \bigg |$

We know,

If x and y are two complex numbers then

$\red{ \boxed{ \sf{ \: |xy| \: = \: |x| \: |y| \: }}}$

So, using this,

$\rm \:  =  \: |x| |y| |z|\bigg |\dfrac{8}{x} + \dfrac{27}{y} + \dfrac{64}{z} \bigg |$

$\rm \:  =  \:2 \times 3 \times 4\bigg |\dfrac{8}{x} + \dfrac{27}{y} + \dfrac{64}{z} \bigg |$

$\red{\bigg \{ \because \: |x| = 2, \: |y| = 3, \: |z| = 4\bigg \}}$

$\rm \:  =  \:24\bigg |\dfrac{8}{x} + \dfrac{27}{y} + \dfrac{64}{z} \bigg |$

Now, we know that

$\red{ \boxed{ \sf{ \:z \bar{z} \: = { |z| }^{2}\bf\implies \:\dfrac{1}{z} = \dfrac{ \bar{z}}{ { |z| }^{2} } }}}$

So, applying this, we get

$\rm \:  =  \:24\bigg |\dfrac{8 \bar{x}}{{ |x| }^{2}} + \dfrac{27\bar{y}}{{ |y| }^{2}} + \dfrac{64\bar{z}}{ { |z| }^{2} } \bigg |$

$\rm \:  =  \:24\bigg |\dfrac{8 \bar{x}}{4} + \dfrac{27\bar{y}}{9} + \dfrac{64\bar{z}}{16} \bigg |$

$\rm \:  =  \: 24 |2\bar{x} + 3\bar{y} + 4\bar{z}|$

We know,

$\red{ \boxed{ \sf{ \:\bar{x} + \bar{y} = \overline{x + y}}}}$

So, using this, we get

$\rm \:  =  \: 24 | \overline{2x + 3y + 4z}|$

$\red{ \boxed{ \sf{ \: |z| = |\bar{z}|}}}$

So, using this, we get

$\rm \:  =  \: 24 |{2x + 3y + 4z}|$

$\red{\bigg \{ \because \: |2x + 3y + 4z| = 4 \bigg \}}$

$\rm \:  =  \: 24 \times 4$

$\rm \:  =  \: 96$