Chapter 6.8, Problem 47WE

a.

To determine

To show $\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ is a square root of $i$

Explanation of Solution

We have to prove $\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i=\sqrt{i}$

Now we square both sides and then solve.

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

Now we solve left hand side of the equation,

We will use the identity ${\left(a+b\right)}^2=a^2+b^2+2ab$

We get,

  $\begin{array}{l}\Rightarrow \left(\frac{2}{4}\right)+\left(\frac{2}{4}{i}^2\right)+i\\ \Rightarrow \frac{2}{4}-\frac{2}{4}+i=i\end{array}$

We got left hand side value’s equal to the right hand side of the equation. Hence we proved that $\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ is a square root of $i$

b.

To determine

To find square root of $i$

Explanation of Solution

Let us assume a number $a+bi$ that is square root of $i$

Now,

  $a+bi=\sqrt{i}$

Squaring both sides we get,

  $\begin{array}{l}{a}^2+b^2{i}^2+2abi=i\\ a^2-b^2+2abi=i\end{array}$

Comparing both sides we get $a^2-b^2=0,2ab=1$

Now we can conclude that $a=\pm b$

Hence we get $b=\pm \frac{1}{\sqrt{2}}=\pm \frac{\sqrt{2}}{2}$

Hence the two roots of $i$ are $\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ and $-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$