A first course in Complex Analysis with Applications

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Assume for the moment that $\sqrt{1+i}$. makes sense in the complex number system.

How would you then demonstrate the validity of the equality $$\sqrt{1+i} = \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{2}} + i\sqrt{-\frac{1}{2} + \frac{1}{2} \sqrt{2}}$$?

Source: Dennis G. Zill & Patrick D. Shanahan

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In order to prove o$$\sqrt{1+i} = \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{2}} + i\sqrt{-\frac{1}{2} + \frac{1}{2} \sqrt{2}}$$

we square the right hand side and show that it is the same as $1+i.$

Note that $$\bigg (\sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{2}} + i\sqrt{-\frac{1}{2} + \frac{1}{2} \sqrt{2}}\bigg )^2 =$$

$$ \bigg( \frac{1}{2} +\frac{1}{2} \sqrt{2}\bigg) -\bigg( \frac{-1}{2} +\frac{1}{2} \sqrt{2}\bigg) + 2i \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{2}} \sqrt{-\frac{1}{2} + \frac{1}{2} \sqrt{2}}=$$

$$ 1 +2i(1/2)=1+i$$