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I am reading the post here Galois Group of $x^{4}+7$ But I cannot see that how can I see the fect that $\sqrt[4]{28} + i\sqrt[4]{28} \;=\; 2\sqrt[4]{-7}$.

May I ask for some steps? Thanks in advance!

EDIT: Instead of a claim-and justify argument. May I ask for **deducing ** from the left hand side?

non-abelian group of order 9
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    Hint: $(1+i)^4=-4$ – guest Jun 01 '17 at 01:24
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    Just raise both sides to the fourth power. – Simply Beautiful Art Jun 01 '17 at 01:24
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    @SimplyBeautifulArt It is not necessarily true that if after raising to some power they are equal implies that they are equal. For instance, take any element in a ring which are both of order $4$. – non-abelian group of order 9 Jun 01 '17 at 01:36
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    *deducing*: $z=\sqrt[4]{28} + i\sqrt[4]{28}=(1+i)\sqrt[4]{28} \implies z^2=2i\sqrt{28}=4i\sqrt{7}=4 \sqrt{-7},$ – dxiv Jun 01 '17 at 01:37
  • How do you define $\sqrt[4]{-7}$? There are four possible values. If you raised both the lhs and the right hand side you get the same value so the lhs is one of the four possible ways to interpret the right hand side. – fleablood Jun 01 '17 at 02:37
  • @non-abeliangroupoforder9 but $\sqrt{-7}$ either isn't well-defined or is a multivalue function. As a multivalued function one of the four values of $2\sqrt{-7}$ is the lhs if raising both sides to the fourth power yield equal results. – fleablood Jun 01 '17 at 02:39

3 Answers3

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$\sqrt[4]{28} + i\sqrt[4]{28} = 2\sqrt[4]{-7}$

$\sqrt{2}\sqrt[4]{7} + i\sqrt{2}\sqrt[4]{7} = 2i^\frac{1}{2}\sqrt[4]{7}$

Remember that $\displaystyle i=\left(\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}\right)^2$

Done.

Saketh Malyala
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Let $a=\sqrt[4]{28}=(2^2\cdot 7)^\frac 14=2^\frac 12\cdot 7^\frac 14$.

$$\begin{align} \sqrt[4]{28} + i\sqrt[4]{28} &=a\cdot (1+i)\\ &=(2^\frac 12\cdot 7^\frac 14)\cdot \sqrt2\ e^{i\pi/4}\\ &=2\cdot (7e^{i\pi})^\frac 14\\ &=2 \cdot (-7)^\frac 14 \end{align}$$

Hypergeometricx
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Note: $\sqrt[4]{-1}=\sqrt{\sqrt{-1}}=\sqrt{i}.$ Hence:

$$\sqrt[4]{28} + i\sqrt[4]{28}=2\sqrt[4]{-7} \Rightarrow $$ $$\sqrt[4]{28}(1 + i)= 2\sqrt[4]{7}\cdot\sqrt[4]{-1} \Rightarrow$$ $$1+i=\sqrt{2}\cdot \sqrt{i} \Rightarrow$$ $$(1+i)^2=2i \Rightarrow$$ $$1+2i+i^2=2i.$$

farruhota
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