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1-Let-z-3-4i-3-4i-where-square-root-is-taken-with-positive-real-part-Then-Re-z-is-a-3-b-4-c-2-d-1-2-Suppose-a-b-c-are-in-AP-and-a-2-b-2-c-2-are-in-GP-If

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Question Number 112364 by john santu last updated on 07/Sep/20
(1) Let z = (√(3+4i)) + (√(−3−4i)) , where square root is taken with positive real part. Then Re(z) is _ (a)3 (b) 4 (c) 2 (d) 1 (2)Suppose a,b,c are in AP and a^2 ,b^2 ,c^2 are in GP. If a<b<c and a+b+c=(3/2), then a = _ (a) (1/(2(√2))) (b) (1/(2(√3))) (c) (((√3)−2)/(2(√3))) (d) (((√2)−2)/(2(√2))) (3)A man sent 7 letters to his 7 friend . The letters are kept in addressed envelopes at random. The probability that 3 friend receive correct letters and 4 letters go to wrong destination is _ (a) (1/8) (b) (1/(16)) (c) (3/(32)) (d) (5/(64))
$$\left(\mathrm{1}\right)\:{Let}\:{z}\:=\:\sqrt{\mathrm{3}+\mathrm{4}{i}}\:+\:\sqrt{−\mathrm{3}−\mathrm{4}{i}}\:,\:{where} \\ $$$${square}\:{root}\:{is}\:{taken}\:{with}\:{positive}\: \\ $$$${real}\:{part}.\:{Then}\:{Re}\left({z}\right)\:{is}\:\_ \\ $$$$\left({a}\right)\mathrm{3}\:\:\:\:\:\left({b}\right)\:\mathrm{4}\:\:\:\:\:\left({c}\right)\:\mathrm{2}\:\:\:\:\:\left({d}\right)\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right){Suppose}\:{a},{b},{c}\:{are}\:{in}\:{AP}\:{and}\:{a}^{\mathrm{2}} ,{b}^{\mathrm{2}} ,{c}^{\mathrm{2}} \\ $$$${are}\:{in}\:{GP}.\:{If}\:{a}<{b}<{c}\:{and}\:{a}+{b}+{c}=\frac{\mathrm{3}}{\mathrm{2}}, \\ $$$${then}\:{a}\:=\:\_ \\ $$$$\left({a}\right)\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\:\:\:\left({b}\right)\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\left({c}\right)\:\frac{\sqrt{\mathrm{3}}−\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\left({d}\right)\:\frac{\sqrt{\mathrm{2}}−\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\left(\mathrm{3}\right){A}\:{man}\:{sent}\:\mathrm{7}\:{letters}\:{to}\:{his}\:\mathrm{7}\:{friend} \\ $$$$.\:{The}\:{letters}\:{are}\:{kept}\:{in}\:{addressed}\: \\ $$$${envelopes}\:{at}\:{random}.\:{The}\:{probability} \\ $$$${that}\:\mathrm{3}\:{friend}\:{receive}\:{correct}\:{letters} \\ $$$${and}\:\mathrm{4}\:{letters}\:{go}\:{to}\:{wrong}\:{destination} \\ $$$${is}\:\_ \\ $$$$\left({a}\right)\:\frac{\mathrm{1}}{\mathrm{8}}\:\:\:\:\:\left({b}\right)\:\frac{\mathrm{1}}{\mathrm{16}}\:\:\:\left({c}\right)\:\frac{\mathrm{3}}{\mathrm{32}}\:\:\:\:\:\left({d}\right)\:\frac{\mathrm{5}}{\mathrm{64}} \\ $$
Answered by MJS_new last updated on 07/Sep/20
(√z) is unique (√(3+4i))=2+i (√(−3−4i))=1−2i ⇒ z=3−i
$$\sqrt{{z}}\:\mathrm{is}\:\mathrm{unique} \\ $$$$\sqrt{\mathrm{3}+\mathrm{4i}}=\mathrm{2}+\mathrm{i} \\ $$$$\sqrt{−\mathrm{3}−\mathrm{4i}}=\mathrm{1}−\mathrm{2i} \\ $$$$\Rightarrow\:{z}=\mathrm{3}−\mathrm{i} \\ $$
Commented by MJS_new last updated on 07/Sep/20
x^2 =4 we′re looking for all possible numbers x which solve the equation ⇒ x=±2 2^2 we just square without caring if or if not (any other number)^2 =2^2 =4 (√x)=2 we′re looking for all possible numbers x which solve the equation ⇒ x=4 (try to find another solution x∈C) (√4) we just extract the root without caring if or if not (√(any other number))=(√4)=2 so there′s no equation given, nothing to solve when we just calculate 7^2 or (√(49)) btw. there′s no solution to (√x)=−4 (try it) z=re^(iθ) with r≥0 and θ∈R z^q =r^q e^(iqθ) nothing to solve with z, q given there′s an exception made to keep geometric problems real: z^(1/n) with z<0 and n=2k+1∧k∈Z ⇒ z=−(−z)^(1/n) or z=re^(iπ) ⇒ z^(1/(2k+1)) =r^(1/(2k+1)) e^(iπ) not r^(1/(2k+1)) e^((iπ)/(2k+1))
$${x}^{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{we}'\mathrm{re}\:\mathrm{looking}\:\mathrm{for}\:{all}\:{possible}\:\mathrm{numbers}\:{x}\:\mathrm{which} \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\Rightarrow\:{x}=\pm\mathrm{2} \\ $$$$\mathrm{2}^{\mathrm{2}} \:\mathrm{we}\:\mathrm{just}\:\mathrm{square}\:\mathrm{without}\:\mathrm{caring}\:\mathrm{if}\:\mathrm{or}\:\mathrm{if}\:\mathrm{not} \\ $$$$\left(\mathrm{any}\:\mathrm{other}\:\mathrm{number}\right)^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$$ \\ $$$$\sqrt{{x}}=\mathrm{2} \\ $$$$\mathrm{we}'\mathrm{re}\:\mathrm{looking}\:\mathrm{for}\:{all}\:{possible}\:\mathrm{numbers}\:{x}\:\mathrm{which} \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\Rightarrow\:{x}=\mathrm{4}\:\left(\mathrm{try}\:\mathrm{to}\:\mathrm{find}\:\mathrm{another}\right. \\ $$$$\left.\mathrm{solution}\:{x}\in\mathbb{C}\right) \\ $$$$\sqrt{\mathrm{4}}\:\mathrm{we}\:\mathrm{just}\:\mathrm{extract}\:\mathrm{the}\:\mathrm{root}\:\mathrm{without}\:\mathrm{caring}\:\mathrm{if} \\ $$$$\mathrm{or}\:\mathrm{if}\:\mathrm{not}\:\sqrt{\mathrm{any}\:\mathrm{other}\:\mathrm{number}}=\sqrt{\mathrm{4}}=\mathrm{2} \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{equation}\:\mathrm{given},\:\mathrm{nothing}\:\mathrm{to} \\ $$$$\mathrm{solve}\:\mathrm{when}\:\mathrm{we}\:\mathrm{just}\:\mathrm{calculate}\:\mathrm{7}^{\mathrm{2}} \:\mathrm{or}\:\sqrt{\mathrm{49}} \\ $$$$ \\ $$$$\mathrm{btw}.\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{to}\:\sqrt{{x}}=−\mathrm{4}\:\left(\mathrm{try}\:\mathrm{it}\right) \\ $$$$ \\ $$$${z}={r}\mathrm{e}^{\mathrm{i}\theta} \:\mathrm{with}\:{r}\geqslant\mathrm{0}\:\mathrm{and}\:\theta\in\mathbb{R} \\ $$$${z}^{{q}} ={r}^{{q}} \mathrm{e}^{\mathrm{i}{q}\theta} \:\mathrm{nothing}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{with}\:{z},\:{q}\:\mathrm{given} \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{an}\:\mathrm{exception}\:\mathrm{made}\:\mathrm{to}\:\mathrm{keep}\:\mathrm{geometric} \\ $$$$\mathrm{problems}\:\mathrm{real}: \\ $$$${z}^{\frac{\mathrm{1}}{{n}}} \:\mathrm{with}\:{z}<\mathrm{0}\:\mathrm{and}\:{n}=\mathrm{2}{k}+\mathrm{1}\wedge{k}\in\mathbb{Z} \\ $$$$\Rightarrow\:{z}=−\left(−{z}\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$\mathrm{or}\:{z}={r}\mathrm{e}^{\mathrm{i}\pi} \:\Rightarrow\:{z}^{\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}} ={r}^{\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}} \mathrm{e}^{\mathrm{i}\pi} \:\boldsymbol{{not}}\:{r}^{\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}} \mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2}{k}+\mathrm{1}}} \\ $$
Commented by john santu last updated on 08/Sep/20
greatt
$${greatt} \\ $$

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