## Trigonometry (11th Edition) Clone

We need to prove $$\Big(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\Big)^2=i$$ first. By doing so, it is true that $$\Big(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\Big)=\sqrt i$$
$$\Big(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\Big)=\sqrt i$$ To prove this, we can prove that $$\Big(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\Big)^2=i$$ We come from the left side first: $$A=\Big(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\Big)^2$$ $$A=\Big(\frac{\sqrt2}{2}\Big)^2+\Big(\frac{\sqrt 2}{2}i\Big)^2+2\times\frac{\sqrt 2}{2}\times\frac{\sqrt 2}{2}i$$ $$A=\frac{1}{2}+\frac{1}{2}i^2+\frac{\sqrt 2\times\sqrt 2}{2}i$$ $$A=\frac{1}{2}+\frac{1}{2}(-1)+\frac{2}{2}i$$ $$A=\frac{1}{2}-\frac{1}{2}+i$$ $$A=i$$ Therefore, it is true that $$\Big(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\Big)^2=i$$ Thus, $$\Big(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\Big)=\sqrt i$$