Why I'm wrong?

Calvin Edit: Discussion about mistakes made in a problem should occur on the solution discussion, so that others who work on the problem can also learn from it. I have locked this discussion.

In Parth Lohomi's question

It is given $\left| { z }_{ 1 } \right| =2,\left| { z }_{ 2 } \right| =3\quad and\quad \left| { z }_{ 3 } \right| =4$ and we have to find the maximum value of $|z_1-z_2|^2 + |z_2-z_3|^2 + |z_3-z_1|^2$ .

here is how I did the problem,

We know that $\left| { z }_{ 1 } \right| =2$ represent a circle with center (0,0) and radius=2. Similarly $\left| { z }_{ 2 } \right| =3$ and $\left| { z }_{ 3 } \right| =4$ represent circles with radius 3 and 4 respectively and centers (0,0)

Now drawing these circles on complex plane we get maximum values of,

$\left| { z }_{ 1 }-{ z }_{ 2 } \right| =5\Longrightarrow { \left| { z }_{ 1 }-{ z }_{ 2 } \right| }^{ 2 }=25$

$\left| { z }_{ 2 }-{ z }_{ 3 } \right| =7\Longrightarrow { \left| { z }_{ 2 }-{ z }_{ 3 } \right| }^{ 2 }=49$

$\left| { z }_{ 3 }-{ z }_{ 1 } \right| =6\Longrightarrow { \left| { z }_{ 3 }-{ z }_{ 1 } \right| }^{ 2 }=36$

Therefore $|z_1-z_2|^2 + |z_2-z_3|^2 + |z_3-z_1|^2$ =25+49+36=110

But this answer is shown wrong......So what has I done wrong in solving this?Please help.

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Calvin Lin Staff - 6 years, 3 months ago

Ok, Thanks sir

Anandhu Raj - 6 years, 3 months ago

@Calvin Lin , @Parth Lohomi @Swananda Punde , , @Deepanshu Gupta, @Ronak Agarwal Could you help please...

Anandhu Raj , it is true that these are the maximum values of the expressions $|z_1 - z_2| \dots$ , but these are not simultaneously satisfied for any triplet. To make my point clearer and simple to understand consider any point $z_1$ and find the corresponding $z_2$ so that $| z_1 - z_2 |$ is $5$ . Similarly, find a $z_3$ for the previously found $z_2$ so that $| z_2 - z_3 | = 7$ . Now you have all the three points $z_1$ , $z_2$ and $z_3$ . Is $| z_1 - z_3 | = 6$ ?

Sudeep Salgia - 6 years, 3 months ago

Oh! got it ..Thanks! So is there any geometrical approach for this question?

@Unnikuttan Cheranalloor The equation ${ (x-3) }^{ 9 }+{ (x-{ 3 }^{ 2 }) }^{ 9 }+{ (x-{ 3 }^{ 3 }) }^{ 9 }+..........+{ (x-{ 3 }^{ 9 }) }^{ 9 }=0$ has

(a) all the roots real.

(b)1 real and 8 imaginary roots.

(c)real roots namely x= $3,{ 3 }^{ 2 },{ 3 }^{ 3 }.....{3}^{9}$

(d)5 real and 4 imaginary roots.

How to do?

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`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$