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Question Number 45450 by Tinkutara last updated on 13/Oct/18

Commented by Meritguide1234 last updated on 14/Oct/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18

	$a_{1}^{2} + b_{1}^{2} = 1 a_{2}^{2} + b_{2}^{2} = 4$ $(a_{1} + {i b}_{1}) (a_{2} + {i b}_{2})$ $a_{1} a_{2} + i (a_{1} b_{2} + a_{2} b_{1}) - b_{1} b_{2}$ $g i v e n (a_{1} a_{2} - b_{1} b_{2}) = 0$ $a_{1} a_{2} = b_{1} b_{2}$ $\frac{a_{1}}{b_{1}} = \frac{b_{2}}{a_{2}} = r$ $a_{1} = b_{1} r b_{2} = a_{2} r$ $a_{1}^{2} + b_{1}^{2} = 1$ $b_{1}^{2} (r^{2} + 1) = 1 a_{2}^{2} (1 + r^{2}) = 4$ $\frac{1}{b_{1}^{2}} = \frac{4}{a_{2}^{2}} a_{2}^{2} = 4 b_{1}^{2}$ $w_{1} = a_{1} + i \frac{a_{2}}{2}$ $∣ w_{1} ∣= \sqrt{a_{1}^{2} + \frac{a_{2}^{2}}{4}}$ $1) ∣ w_{1} ∣= \sqrt{a_{1}^{2} + b_{1}^{2}} = 1 \leftarrow a n s w e r o f q u e s t i o n 1)$ $w_{2} = 2 b_{1} + {i b}_{2}$ $2) ∣ w_{2} ∣= \sqrt{4 b_{1}^{2} + b_{2}^{2}} = \sqrt{a_{2}^{2} + b_{2}^{2}} = \sqrt{4} = 2 \leftarrow a n s o f q 2)$ $w_{1} w_{2} = (a_{1} + i \frac{a_{2}}{2}) (2 b_{1} + {i b}_{2})$ $= 2 a_{1} b_{1} + i (a_{2} b_{1} + a_{1} b_{2}) - \frac{a_{2} b_{2}}{2}$ $R e (w_{1} w_{2}) = (2 a_{1} b_{1} - \frac{a_{2} b_{2}}{2})$ $a_{1}^{2} + b_{1}^{2} = 1 a_{2}^{2} + b_{2}^{2} = 4$ $\frac{a_{1}}{b_{1}} = \frac{b_{2}}{a_{2}}$ $\frac{a_{1}}{b_{1}} = \frac{b_{2}}{2 b_{1}} b_{2} = 2 a_{1} a_{2} = 2 b_{1}$ $s o R e (w_{1} w_{2}) = (2 a_{1} b_{1} - \frac{a_{2} b_{2}}{2})$ $= (a_{1} a_{2} - a_{1} a_{2}) = 0 \leftarrow a n s q 3$ $4) i m (w_{1} w_{2}) = (a_{2} b_{1} + a_{1} b_{2})$ $= (2 b_{1}^{2} + 2 a_{1}^{2})$ $= 2 (a_{1}^{2} + b_{1}^{2}) = 2 \times 1 = 2$ $s o i m (w_{1} w_{2}) n o t e q u s l s t o z e r o$ $i m (w_{1} w_{2}) = 2 \times 1 = 2$
	Commented by Tinkutara last updated on 13/Oct/18
	Thank you very much Sir! I got the answer. ��
	Commented by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18

	$t h a n k y o u a l s o f o r c r e a t i n g s u c h p l a t f o r m$ $t o e n g a g e o u r m i n d . .$
	Commented by Tinkutara last updated on 13/Oct/18
	Sorry Sir, but I am not the developer of this app. I am just a user like you.
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