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Question Number 51389 by Tawa1 last updated on 26/Dec/18

Commented by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18

excellent problem...give time to find the way to reach goal...

$${excellent}\:{problem}...{give}\:{time}\:{to}\:{find}\:{the}\:{way} \\ $$$${to}\:{reach}\:{goal}... \\ $$

Commented by Tawa1 last updated on 26/Dec/18

Alright sir, i will wait. God bless you sir

$$\mathrm{Alright}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{will}\:\mathrm{wait}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by Tinkutara last updated on 26/Dec/18

Commented by Tawa1 last updated on 26/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by Tawa1 last updated on 26/Dec/18

But sir, i don′t really understand the steps. ∣a∣^2 = ((√(z_1 ^2 + z_2 ^2 + z_3 ^2 )))^2 i think. please let me understand

$$\mathrm{But}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{really}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{steps}.\:\: \\ $$$$\mid\mathrm{a}\mid^{\mathrm{2}} \:\:=\:\:\left(\sqrt{\mathrm{z}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{z}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{z}_{\mathrm{3}} ^{\mathrm{2}} }\right)^{\mathrm{2}} \:\:\mathrm{i}\:\mathrm{think}.\:\:\mathrm{please}\:\mathrm{let}\:\mathrm{me}\:\mathrm{understand} \\ $$

Commented by Tawa1 last updated on 27/Dec/18

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Tinkutara last updated on 27/Dec/18

No, ∣a∣=a×a^− Since zz^− =∣z∣^2

$${No},\:\mid{a}\mid={a}×\overset{−} {{a}} \\ $$$${Since}\:{z}\overset{−} {{z}}=\mid{z}\mid^{\mathrm{2}} \\ $$

Commented by Tawa1 last updated on 28/Dec/18

Sir, why the ω interchanged in ∣b_1 ∣ and ∣c_1 ∣ ??. wz_2 ...... w^2 z_2 ^− ..... why the change.

$$\mathrm{Sir},\:\mathrm{why}\:\mathrm{the}\:\omega\:\mathrm{interchanged}\:\mathrm{in}\:\mid\mathrm{b}_{\mathrm{1}} \mid\:\mathrm{and}\:\mid\mathrm{c}_{\mathrm{1}} \mid\:??. \\ $$$$\:\:\mathrm{wz}_{\mathrm{2}} \:\:\:\:......\:\:\:\:\:\mathrm{w}^{\mathrm{2}} \:\:\overset{−} {\mathrm{z}}_{\mathrm{2}} \:\:.....\:\:\:\:\:\mathrm{why}\:\mathrm{the}\:\mathrm{change}.\:\: \\ $$

Commented by Tawa1 last updated on 27/Dec/18

Sir, i am trying to understand the expansion. I don′t understand please

$$\mathrm{Sir},\:\mathrm{i}\:\mathrm{am}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{expansion}. \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{please} \\ $$

Commented by Tinkutara last updated on 27/Dec/18

I have left some terms there. If you write them, you can see they also add up to 0. What you didn't understand?

Commented by Tinkutara last updated on 28/Dec/18

Because ω^− =ω^2

$${Because}\:\overset{−} {\omega}=\omega^{\mathrm{2}} \\ $$

Commented by Tawa1 last updated on 06/Jan/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by Tawa1 last updated on 06/Jan/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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