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Question Number 112885 by malwan last updated on 10/Sep/20

Commented by malwan last updated on 10/Sep/20

$$P=?$$$$\left(a\right)(-5\pi /4,2^{1/2})$$$$\left(b\right)(-5\pi /4,2^{-1/2})$$$$\left(c\right)(-7\pi /4,2^{1/2})$$$$\left(d\right)(-7\pi /4,2^{-1/2})$$

Commented by malwan last updated on 10/Sep/20

$$Q=?$$$$\left(a\right)(\frac{5\pi }{4},-2^{\frac{1}{2}})$$$$\left(b\right)(\frac{5\pi }{4},-2^{-\frac{1}{2}})$$$$\left(c\right)(\frac{7\pi }{4},-2^{\frac{1}{2}})$$$$\left(d\right)(\frac{7\pi }{4},-2^{-\frac{1}{2}})$$

Commented by Aziztisffola last updated on 10/Sep/20

$$\left(d\right)\mathrm{P}(-7\pi /4,2^{-1/2})$$

Commented by Aziztisffola last updated on 10/Sep/20

$$\left(b\right)\mathrm{Q}(\frac{5\pi }{4},-2^{-\frac{1}{2}})$$

Commented by malwan last updated on 10/Sep/20

$$thankyou$$

Commented by malwan last updated on 10/Sep/20

$$but-2^{-\frac{1}{2}}=-\frac{1}{\sqrt{-2}}???$$

Commented by Aziztisffola last updated on 10/Sep/20

$$-2^{-\frac{1}{2}}=-\frac{1}{2^{\frac{1}{2}}}=-\frac{1}{\sqrt{2}}$$$$$$

Commented by malwan last updated on 10/Sep/20

$$sorry$$$$Ithought-2^{-\frac{1}{2}}={(-2)}^{-\frac{1}{2}}$$$$thankyousir$$

Commented by Aziztisffola last updated on 10/Sep/20

$$-2^{-\frac{1}{2}}\ne {(-2)}^{-\frac{1}{2}}$$$$\mathrm{and}{(-2)}^{-\frac{1}{2}}\mathrm{is}\mathrm{not}\mathrm{define}\mathrm{in}\mathbb{R}.$$

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