Complex Complex Numbers !!

Algebra Level 2

$|Z_{1}| = |Z_{2}| = |Z_{3}| = \Bigg|\dfrac{1}{Z_1} + \dfrac{1}{Z_{2}} + \dfrac{1}{Z_{3}} \Bigg| = 1$

If $Z_{1} , Z_{2} , Z_{3}$ are complex numbers in the argand plane satisfying the above equalities, find the value of:

$\bigg|Z_{1} + Z_{2} + Z_{3} \bigg| = \ ?$

The answer is 1.

4 solutions

A Former Brilliant Member
Dec 23, 2014

$|{ Z }_{ 1 }|\quad =\quad |{ Z }_{ 2 }|\quad =\quad |{ Z }_{ 3 }|\quad =\quad 1\\ \\ As\quad { |Z| }^{ 2 }\quad =\quad Z\quad \times \quad \bar { Z } \quad =\quad 1,\quad where\quad \bar { Z } \quad is\quad conjugate\quad of\\ \\ Z\\ \\ So,\quad \bar { { Z }_{ 1 } } \quad =\quad \frac { 1 }{ { Z }_{ 1 } } ;\quad \bar { { Z }_{ 2 } } \quad =\quad \frac { 1 }{ { Z }_{ 2 } } ;\quad \bar { { Z }_{ 3 } } \quad =\quad \frac { 1 }{ { Z }_{ 3 } } \quad ----1.\\ \\ Therefore,\quad |\frac { 1 }{ { Z }_{ 1 } } +\frac { 1 }{ { Z }_{ 2 } } +\frac { 1 }{ { Z }_{ 3 } } |\quad =\quad 1\\ \\ From\quad 1;\\ |\bar { { Z }_{ 1 } } +\bar { { Z }_{ 2 } } +\bar { { Z }_{ 3 } } |\quad =\quad 1\\ Now,\quad remember\quad this\quad equation.\\ \\ Here\quad comes\quad the\quad tricky\quad part,\\ \\ Let\quad Re({ Z }_{ 1 }+{ Z }_{ 2 }+{ Z }_{ 3 })\quad =\quad X\\ And\quad Im({ Z }_{ 1 }+{ Z }_{ 2 }+{ Z }_{ 3 })\quad =\quad Y\\ \\ Therefore\quad Re(\bar { { Z }_{ 1 } } +\bar { { Z }_{ 2 } } +\bar { { Z }_{ 3 } } )\quad =\quad X\\ But,\quad Im(\bar { { Z }_{ 1 } } +\bar { { Z }_{ 2 } } +\bar { { Z }_{ 3 } } )\quad =\quad -Y\\ \\ Clearly,\quad |{ Z }_{ 1 }+{ Z }_{ 2 }+{ Z }_{ 3 }|\quad =\quad \sqrt { { X }^{ 2 }+{ Y }^{ 2 } } \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \sqrt { { X }^{ 2 }+{ (-Y) }^{ 2 } } \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad |\bar { { Z }_{ 1 } } +\bar { { Z }_{ 2 } } +\bar { { Z }_{ 3 } } |\\ But\quad |\bar { { Z }_{ 1 } } +\bar { { Z }_{ 2 } } +\bar { { Z }_{ 3 } } |\quad =\quad 1\\ So,\quad |{ Z }_{ 1 }+{ Z }_{ 2 }+{ Z }_{ 3 }|\quad =\quad 1\\$

CHEERS!!:):)

I don't think $|Z| = Z \times \bar{Z}$ is correct. Rather $|Z|^{2} = Z \times \bar{Z}$ is correct . So your first step itself is not correct, check it again mate !! However it doesn't affect the solution because $1^{2} = 1$

Ayush Parasar - 6 years, 5 months ago

Ooh!! Sorry mate...and thanks...i forgot to mention that !:)

A Former Brilliant Member - 6 years, 5 months ago

Chew-Seong Cheong
Jan 3, 2015

Let $z_n=a_n+b_ni$ , where $n=1,2,3$ .

Then $|z_n| = 1 \quad \Rightarrow \sqrt{a_n^2 + b_n^2} = a_n^2 + b_n^2 = 1$

Now, $\dfrac {1}{z_n} = \dfrac {1}{a_n+b_ni} = \dfrac {a_n-b_ni}{(a_n+b_ni)(a_n-b_ni)}$

$\quad \quad \quad \quad = \dfrac {a_n-b_ni}{a_n^2 + b_n^2} = \dfrac {a_n-b_ni}{1} = a_n-b_ni$

$\Rightarrow \dfrac {1}{z_1}+\dfrac {1}{z_2}+\dfrac {1}{z_3} = a_1-b_1i + a_2-b_2i + a_3-b_3i$

$\quad \quad \quad \quad \quad \quad \quad \quad = (a_1 + a_2 + a_3) - (b_1 + b_2 + b_3)i$

Now $\left|\dfrac {1}{z_1}+\dfrac {1}{z_2}+\dfrac {1}{z_3} \right| = 1$

$\Rightarrow \sqrt{(a_1 + a_2 + a_3)^2 + (b_1 + b_2 + b_3)^2} = 1$

Also $|z_1+z_2+z_3| = \sqrt{(a_1 + a_2 + a_3)^2 + (b_1 + b_2 + b_3)^2} = \boxed{1}$

Sir Can you please elaborate the last line: | $z_{1}+z_{2}+z$ + 3 |

Because I am confuse about 3 and z

Syed Baqir - 5 years, 8 months ago

Sorry, it was a typo. It should be $|z_1+z_2+z_3|$ .

Chew-Seong Cheong - 5 years, 8 months ago

ahh, Thanks .

SIr, if we expand (a1 + a2 + a3 )^2 + (b1 + b2 + b3 ) ^2

will it give us Z1 , Z2 and Z3 respectively ?

Syed Baqir - 5 years, 8 months ago

@Syed Baqir – No, $z_1$ , $z_2$ and $z_3$ are complex numbers. They are with imaginary part. $z_n = a_n + b_n \color{#D61F06}{i}$ ; $\color{#D61F06}{i} = \sqrt{-1}$ is the imaginary unit. $a_n$ and $b_n$ are real numbers.