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Question-112885

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Question Number 112885 by malwan last updated on 10/Sep/20
Commented by malwan last updated on 10/Sep/20
P=? (a) (−5π/4,2^(1/2) ) (b) (−5π/4,2^(−1/2) ) (c) (−7π/4,2^(1/2) ) (d) (−7π/4,2^(−1/2) )
$${P}=? \\ $$$$\left({a}\right)\:\left(−\mathrm{5}\pi/\mathrm{4},\mathrm{2}^{\mathrm{1}/\mathrm{2}} \right) \\ $$$$\left({b}\right)\:\left(−\mathrm{5}\pi/\mathrm{4},\mathrm{2}^{−\mathrm{1}/\mathrm{2}} \right) \\ $$$$\left({c}\right)\:\left(−\mathrm{7}\pi/\mathrm{4},\mathrm{2}^{\mathrm{1}/\mathrm{2}} \right) \\ $$$$\left({d}\right)\:\left(−\mathrm{7}\pi/\mathrm{4},\mathrm{2}^{−\mathrm{1}/\mathrm{2}} \right) \\ $$
Commented by malwan last updated on 10/Sep/20
Q=? (a) (((5π)/4) , −2^(1/2) ) (b) (((5π)/4) , −2^(−(1/2)) ) (c) (((7π)/4) , −2^(1/2) ) (d) (((7π)/4) , −2^(−(1/2)) )
$${Q}=? \\ $$$$\left({a}\right)\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}}\:,\:−\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \right) \\ $$$$\left({b}\right)\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}}\:,\:−\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right) \\ $$$$\left({c}\right)\:\left(\frac{\mathrm{7}\pi}{\mathrm{4}}\:,\:−\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \right) \\ $$$$\left({d}\right)\:\left(\frac{\mathrm{7}\pi}{\mathrm{4}}\:,\:−\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right) \\ $$
Commented by Aziztisffola last updated on 10/Sep/20
(d) P(−7π/4,2^(−1/2) )
$$\left({d}\right)\:\mathrm{P}\left(−\mathrm{7}\pi/\mathrm{4},\mathrm{2}^{−\mathrm{1}/\mathrm{2}} \right) \\ $$
Commented by Aziztisffola last updated on 10/Sep/20
(b) Q(((5π)/4) , −2^(−(1/2)) )
$$\left({b}\right)\:\mathrm{Q}\left(\frac{\mathrm{5}\pi}{\mathrm{4}}\:,\:−\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right) \\ $$
Commented by malwan last updated on 10/Sep/20
thank you
$${thank}\:{you} \\ $$
Commented by malwan last updated on 10/Sep/20
but −2^(−(1/2)) = − (1/( (√(−2)))) ???
$${but}\:−\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:−\:\frac{\mathrm{1}}{\:\sqrt{−\mathrm{2}}}\:??? \\ $$
Commented by Aziztisffola last updated on 10/Sep/20
−2^(−(1/2)) =−(1/2^(1/2) )=−(1/( (√2)))
$$−\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{2}}} =−\frac{\mathrm{1}}{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} }=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$ \\ $$
Commented by malwan last updated on 10/Sep/20
sorry I thought −2^(−(1/2)) = (−2)^(−(1/2)) thank you sir
$${sorry} \\ $$$${I}\:{thought}\:−\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\left(−\mathrm{2}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${thank}\:{you}\:{sir} \\ $$
Commented by Aziztisffola last updated on 10/Sep/20
−2^(−(1/2)) ≠ (−2)^(−(1/2)) and (−2)^(−(1/2)) is not define in R.
$$−\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\neq\:\left(−\mathrm{2}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{and}\:\left(−\mathrm{2}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{is}\:\mathrm{not}\:\mathrm{define}\:\mathrm{in}\:\mathbb{R}. \\ $$

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