Question-184481

Question Number 184481 by yaslm last updated on 07/Jan/23

Answered by mahdipoor last updated on 07/Jan/23

i = e^{0} (c o s \frac{π}{2} + i s i n \frac{π}{2}) = e^{\frac{π}{2} i} \Rightarrow

Z_{1} = {(i^{\frac{100}{i}})}^{0.5} = {(e^{(\frac{π}{2} i) (\frac{100}{i})})}^{0.5} = {(e^{\frac{100 π}{2}})}^{0.5} = {(e^{50 π})}^{0.5}

Z_{1} = \pm e^{25 π}

Z_{2} = e^{\frac{π \sqrt{3}}{2} i} = i s i n (\frac{π \sqrt{3}}{2})

Z_{3} = e^{\frac{9 π}{4} i} = i s i n (\frac{9 π}{4})

Commented by yaslm last updated on 07/Jan/23

thanks complete sir

Commented by Frix last updated on 07/Jan/23

If {(e^{50 π})}^{0.5} = \pm e^{25 π} then why e^{\frac{π}{2} i} which can

be written as {(e^{π i})}^{0.5} \neq \pm e^{\frac{π}{2} i} ? Same for e^{\frac{π \sqrt{3}}{2} i} .

Even worse : e^{\frac{9 π}{4} i} = {(e^{9 π i})}^{0.25} = {\begin{cases} \pm e^{\frac{9 π}{4} i} \\ \pm {ie}^{\frac{9 π}{4} i} \end{cases}

{You}^{'} re not using the same rule in each case \dots

Answered by Frix last updated on 07/Jan/23

z = r e^{i θ}; r =∣ z ∣ \Rightarrow r \in R_{0}^{+}; - π < θ ⩽ π

z^{c} = r^{c} e^{i c θ} \forall c \in Z which is always unique

Z_{1} = \sqrt{i^{- 100 i}} = \sqrt{{(e^{i \frac{π}{2}})}^{- 100 i}} = \sqrt{e^{- 50 π i^{2}}} = \sqrt{e^{50 π}} = e^{25 π}

Z_{2} = i^{\sqrt{3}} = {(e^{i \frac{π}{2}})}^{\sqrt{3}} = e^{i \frac{π \sqrt{3}}{2}} = \cos \frac{π \sqrt{3}}{2} + i \sin \frac{π \sqrt{3}}{2}

Z_{3} = i^{\frac{9}{2}} = {(e^{i \frac{π}{2}})}^{\frac{9}{2}} = e^{i \frac{9 π}{4}} = e^{i \frac{π}{4}} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i

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