Q14P - Mathematical Methods in Physical Sciences

Question

Follow steps (a), (b), (c) above to find all the values of the indicated roots $\sqrt[\mathbf{4}]{\mathbf{-}\mathbf{1}}$.

Step-by-Step Solution

Answer

The value of $\sqrt[4]{-1} is \frac{\left(\pm 1\pm i\right)}{\sqrt{2}}$.

The graph used in this question to find the answer is shown below.

Step 1: Given Information

The given expression is $\sqrt[4]{-1}$.

Step 2: Definition of Complex Number

Complex numbers consist of real numbers and imaginary numbers; a complex  can be written in the form of: 

z=a+ib

 

Here a and b are real numbers, and i is the imaginary number which is known as iota, whose value is $\sqrt{-1}$ .

Step 3: Find the value of r and

The Complex number is in the form $-1+0\mathrm{i}$.

 

x=-1

y=0 

The polar coordinates of the point are in the form of $z=r{e}^{i\theta }$.

 

$$\begin{gathered} r=1 \\ \theta =\pi , or 3\pi ,5\pi ,7\pi ,9\pi ,..... \end{gathered}$$

 

The equation $z=r{e}^{i\theta }$ can also be written in another form.

 

$$\begin{gathered} z^{\frac{1}{n}}={\left(r{e}^{i\theta }\right)}^{\frac{1}{n}} \\ =r^{\frac{1}{n}}{e}^{\frac{i\theta }{n}} \\ {z}^{\frac{1}{n}}=\sqrt[n]{r}\left(\cos \frac{\theta }{n}+i \sin \frac{\theta }{n},\right).......\left(1\right) \end{gathered}$$

 

When n=4 , the equation becomes the 4th root of the complex number.

$$\begin{gathered} z^{\frac{1}{4}}=r^{\frac{1}{4}}{e}^{\frac{i\theta }{4}} \\ r=1 \\ \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4},\frac{9\pi }{4},....... \end{gathered}$$ 

Step 4: Plotting of the polar coordinate points on the graph.

It is clear from the above graph that the points $\left(1,\frac{\pi }{4}\right)$ and the point $\left(1,\frac{9\pi }{4}\right)$are the same.

The radius of the circle is 1 and equally spaced $\frac{\pi }{2}$ apart.

$$\begin{gathered} r=1 \\ \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4} \end{gathered}$$

 

Hence, the final answer is $\sqrt[4]{-1}=\frac{\left(\pm 1\pm i\right)}{\sqrt{2}}$.