my answer would be $i$ because I used the definition $i=\sqrt-1$ and replaced the $i$ with that
I'm not sure if I got it right though
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6Do you mean $(\sqrt{i})^i$ or $\sqrt{i^i}$? – Wojowu Mar 08 '19 at 15:22
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@Wojowu in either case, the result is the same. – Decaf-Math Mar 08 '19 at 15:26
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Try considering taking logarithm on both sides of $y=(\sqrt{i})^i$. – Yadati Kiran Mar 08 '19 at 15:27
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2You can see $i^i$. – Mauro ALLEGRANZA Mar 08 '19 at 15:27
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You can see square roots of complex numbers here, in particular $\sqrt{i^i}$. – Dietrich Burde Mar 08 '19 at 15:28
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Another references for $i^i$ on this site is here. – Dietrich Burde Mar 08 '19 at 15:31
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$$\sqrt{i}^i=i^{i/2}=\exp \left(\left(\dfrac{\pi i}{2}+2\pi i n\right)\cdot \dfrac{i}{2}\right)=\exp \dfrac{-\pi}{4}\cdot\exp-\pi n, \ n\in \mathbb{Z}$$
Paras Khosla
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Complex exponentiation doesn't work like real exponentiation. There are infinitely many values of $i^i$ (and two square roots of each.) – saulspatz Mar 08 '19 at 15:36
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This uses one of the values of $\log i$. There are many more, which will result in a different value – Ross Millikan Mar 08 '19 at 15:37
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@RossMillikan Would adding $2\pi n, \ n\in \mathbb{Z}$ in the expression for $i$ generalize? – Paras Khosla Mar 08 '19 at 15:38
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How about $i^{i/2}=\exp\left(\frac{i\pi}{2}\cdot \frac{i}{2}+2\pi ik\cdot\frac{i}{2} \right)=\exp\left(-\left(\frac{\pi}{4}+\pi k\right)\right)$ for $k\in\mathbb{Z}$ – Dispersion Mar 08 '19 at 15:41