Question

Find all the complex roots. Leave your answers in polar form with the argument in degrees.
14) The complex fourth roots of -16

Answer 1

Answer to Question #88644 – Math – Trigonometry

Question

Find all the complex roots. Leave your answers in polar form with the argument in degrees.

14) The complex fourth roots of -16

Solution

The generalized expression of a complex number in polar coordinate is written as:

$$z = r e^{i\theta} = r \times (\cos \theta + i \sin \theta) \text{ Where } \theta = \arg z$$

Now, we can rewrite the above complex number expression into more generalized polar form of complex number as:

$$z = r \times e^{(i\theta + 2n\pi i)} = r \times e^{\left\{i \times (\theta + 2n\pi)\right\}} = r \times e^{\left\{i \times \left(\theta + n \times 360{}^{\circ}\right)\right\}}, \text{ Where } n = 0, \pm 1, \pm 2, \dots$$

Now, we have to obtain complex fourth root of -16. So, mathematically, we have to obtain

$$(-16)^{\frac{1}{4}}. \text{ Let us assume that: } z = (-16)^{\frac{1}{4}} \Rightarrow z^4 = -16$$

So, we first express this expression $z^4 = -16$, into polar coordinate. This is as below:

$$\begin{array}{l} z^4 = -16 \\ \Rightarrow z^4 = 16 \times -1 \\ \Rightarrow z^4 = 16 \times \left\{1 \times \left(\cos 180{}^{\circ} + \sin 180{}^{\circ}\right)\right\} \\ \Rightarrow z^4 = 16 \times e^{(i \times 180{}^{\circ})} \end{array}$$

So, we can rewrite this expression into more generalized polar form of complex number as:

$$\begin{array}{l} z^4 = 16 \times e^{(i \times 180{}^{\circ})} \quad \text{ Where } n = 0, \pm 1, \pm 2, \dots \\ \Rightarrow z^4 = 16 \times e^{i(180{}^{\circ} + n \times 360{}^{\circ})} \end{array}$$

Now, De Moivre’s Theorem to obtain the $k$-th root of a complex number, is written as:

$$\left(e^{i\theta}\right)^{\frac{1}{k}} = \left(\cos \theta + i \sin \theta\right)^{\frac{1}{k}} = \left[ \cos \left\{\frac{\left(\theta + n \times 360{}^{\circ}\right)}{k} \right\} + \sin \left\{\frac{\left(\theta + n \times 360{}^{\circ}\right)}{k} \right\} \right]$$

Where $n = 0, \pm 1, \pm 2, \dots$

So, using this De Moivre’s Theorem, the 4-th root of the given complex number becomes:

$$\begin{array}{l} z^{4} = 16 \times e^{i \left(180{}^{\circ} + n \times 360{}^{\circ}\right)} \\ \Rightarrow z = (16)^{\frac{1}{4}} \times e^{\left(\left\{i \times \left(180{}^{\circ} + n \times 360{}^{\circ}\right)\right\} / 4\right)} \\ \Rightarrow z = 2 \times e^{\left(\left\{i \times \left(180{}^{\circ} + n \times 360{}^{\circ}\right)\right\} / 4\right)} \end{array}$$

So, now putting the values of $n = 0, \pm 1, \pm 2, \ldots$, we can get individual roots of the given number. So, rewriting the expression again, we get:

$$(-16)^{\frac{1}{4}} = 2 \times e^{\left(\left\{i \times \left(180{}^{\circ} + n \times 360{}^{\circ}\right)\right\} / 4\right)} \text{ Where } n = 0, \pm 1, \pm 2, \dots$$

So, we can write the general expression of root for this complex number as:

$$z = (-16)^{\frac{1}{4}} = 2 \times e^{\left(\left\{i \times \left(180{}^{\circ} + n \times 360{}^{\circ}\right)\right\} / 4\right)} \quad \text{ Where } n = 0, \pm 1, \pm 2, \dots$$

Now, to get the distinct roots, we have to put the values of:

$$n = 0, 1, 2, 3$$

And accordingly, the individual root becomes:

**First Root (z₁) for $n = 0$:**

$$z_{1} = 2 \times e^{\left(\left\{i \times \left(180{}^{\circ} + 0 \times 360{}^{\circ}\right)\right\} / 4\right)} = 2 e^{i \times 45{}^{\circ}} = 2 \left(\cos 45{}^{\circ} + i \sin 45{}^{\circ}\right)$$

**Second Root (z₂) for $n = 1$:**

$$\begin{array}{l} z_{2} = 2 \times e^{\left(\left\{i \times \left(180{}^{\circ} + 1 \times 360{}^{\circ}\right)\right\} / 4\right)} \\ \Rightarrow z_{2} = 2 e^{i \times \left(\frac{540{}^{\circ}}{4}\right)} \\ \Rightarrow z_{2} = 2 e^{i \times 135{}^{\circ}} \\ \Rightarrow z_{2} = 2 \left(\cos 135{}^{\circ} + i \sin 135{}^{\circ}\right) \\ \Rightarrow z_{2} = 2 \left\{\cos (90 + 45){}^{\circ} + i \sin (90 + 45){}^{\circ} \right\} \\ \Rightarrow z_{2} = 2 \left(-\sin 45{}^{\circ} + i \cos 45{}^{\circ}\right) \\ \end{array}$$

**Third Root (z₃) for $n = 2$:**

$$\begin{array}{l} z_{3} = 2 \times e^{\left( \frac{ \left\{ i \times \left(180{}^\circ + 2 \times 360{}^\circ\right) \right\} }{4} \right) } \\ \Rightarrow z_{3} = 2 e^{i \times \left( \frac{900{}^\circ}{4} \right) } \\ \Rightarrow z_{3} = 2 e^{i \times 225{}^\circ} \\ \Rightarrow z_{3} = 2 \left( \cos 225{}^\circ + i \sin 225{}^\circ \right) \\ \Rightarrow z_{3} = 2 \left\{ \cos (180 + 45){}^\circ + i \sin (180 + 45){}^\circ \right\} \\ \Rightarrow z_{3} = 2 \left( -\cos 45{}^\circ - i \sin 45{}^\circ \right) \\ \end{array}$$

And finally the fourth Root $(z_4)$ for $n = 3$:

$$\begin{array}{l} z_{4} = 2 \times e^{\left( \frac{ \left\{ i \times \left(180{}^\circ + 3 \times 360{}^\circ\right) \right\} }{4} \right) } \\ \Rightarrow z_{4} = 2 e^{i \times \left( \frac{1260{}^\circ}{4} \right) } \\ \Rightarrow z_{4} = 2 e^{i \times 315{}^\circ} \\ \Rightarrow z_{4} = 2 \left( \cos 315{}^\circ + i \sin 315{}^\circ \right) \\ \Rightarrow z_{4} = 2 \left\{ \cos (360 - 45){}^\circ + i \sin (360 - 45){}^\circ \right\} \\ \Rightarrow z_{4} = 2 \left( \cos 45{}^\circ - i \sin 45{}^\circ \right) \\ \end{array}$$

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