4th roots of complex number 7 + 24i

I was solving a complex numbers worksheet and came across a 4th root question of the complex number $$z = 7 + 24i$$, with the answer in cartesian form. I got that the modulus was 25, and the argument was $$\arctan(\frac{24}{7}),$$ but when solving with these, my answer it differed from the format of the actual answer.

In polar form, my complex number was written as follows: $$25 \times cis(73.3)$$

I'll use one of my roots as an example of the difference: $$2.121+0.692i.$$

I used the rule $$z_k = \sqrt[n]r \left[\cos\frac{\theta + 360k}n + i \sin\frac{\theta + 360k}n\right], k = 0, 1, 2, ... ,(n-1)$$ to solve this question.

Question (part b)

• Please use MathJax. Commented May 2 at 20:24
• You use $z=7+24i$ and later write $z=...$ for the example fourth root. Don't reuse variables like that. Commented May 2 at 20:25
• Also, when we do math here, by default, we do not use degrees here, unless you explicitly say so. Replace $360$ with $2\pi$ or state somewhere your angles are in degrees. Commented May 2 at 20:32
• So ... is this right... $$\arctan\frac{24}{7} =4\arctan\frac13$$ Commented May 2 at 20:32
• $\arctan(24/7)$ is closer to $73.7^\circ$ than to $73.3^\circ$ Commented May 2 at 21:28

take square root twice....First, we want real $$x,y$$ such that $$(x+iy)^2 = 7 + 24 i,$$ or $$x^2 - y^2 = 7 \; , \; \; \; 2xy = 24$$

The latter gives us $$y = \frac{12}{x},$$ so the first becomes $$x^2 - \frac{144}{x^2} = 7,$$ $$x^4 - 144 = 7 x^2$$ $$x^4 - 7 x^2 - 144 = 0$$ From the quadratic formula we get $$x^2 = \frac{7 \pm \sqrt{49 + 576}}{2} = \frac{7 \pm \sqrt{625}}{2} = \frac{7 \pm 25}{2} = \frac{32 \; , \; -18}{2} = 16 \; , \; -9$$ So $$x^2$$ is either $$16$$ or $$-9,$$ and we demanded $$x$$ real, so $$x = 4$$ from which $$2xy = 24$$ says $$y=3$$ So, that is one square root, $$4+3i$$ while the second is $$-4-3i$$

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Starting over, second square root of $$4 + 3 i$$ $$x^2 - y^2 = 4 \; , \; \; \; 2xy = 3$$

• Totally forgot about that way, thanks a lot man. Commented May 3 at 20:56

Let $$z = r \operatorname{cis}(\theta)$$ be one of the roots. Then, by definition of fourth root:

$$(r \operatorname{cis}(\theta))^4 = 7 + 24i$$

Taking the absolute value gives:

$$r^4 = 25$$ $$r = \sqrt{5}$$

So now, we can just divide by $$r^4 = 25$$ to get an equation for $$\theta$$.

$$\operatorname{cis}(\theta)^4 = \frac{7}{25} + \frac{24}{25}i$$

Now, let's expand $$\operatorname{cis}(\theta)$$ to $$c + is$$, where $$c = \cos(\theta)$$ and $$s = \sin(\theta)$$. Then:

$$(c+is)^4 = \frac{7}{25} + \frac{24}{25}i$$ $$c^4 + 4ic^3s - 6c^2s^2 - 4ics^3 + s^4 = \frac{7}{25} + \frac{24}{25}i$$

Breaking this down into real and imaginary parts gives:

$$c^4 - 6c^2s^2 + s^4 = \frac{7}{25} \tag{1}$$ $$4c^3s - 4cs^3 = \frac{24}{25} \tag{2}$$

We can simplify equation (1) by using the Pythagorean identity $$c^2 + s^2 = 1$$ and replace $$c^2 \to 1 - s^2$$.

$$(1-s^2)^2 - 6(1-s^2)s^2 + s^4 = \frac{7}{25}$$ $$1 - 2s^2 + s^4 - 6s^2 + 6s^4 + s^4 = \frac{7}{25}$$ $$8s^4 - 8s^2 + 1 - \frac{7}{25} = 0$$ $$200s^4 - 200s^2 + 18 = 0$$ $$100s^4 - 100s^2 + 9 = 0$$

Using the quadratic formula to solve for $$s^2$$ gives:

$$s^2 = \frac{-(-100) \pm \sqrt{(-100)^2-4(100)(9)}}{2(100)}$$ $$s^2 = \frac{100 \pm \sqrt{6400}}{200}$$ $$s^2 = \frac{100 \pm 80}{200}$$ $$s^2 = \frac{9}{10} \text{ or } s^2 = \frac{1}{10}$$ $$s = \pm\frac{3}{\sqrt{10}} \text{ or } s = \pm\frac{1}{\sqrt{10}}$$

But since $$c^2 = 1 - s^2$$, then:

$$c^2 = \frac{1}{10} \text{ or } c^2 = \frac{9}{10}$$ $$c = \pm \frac{1}{\sqrt{10}} \text{ or } c = \pm \frac{3}{\sqrt{10}}$$

So, taking all the combinations of $$c$$ and $$s$$ gives 8 potential solutions:

$$c + is \in \{ \frac{1}{\sqrt{10}} + i \frac{3}{\sqrt{10}}, \frac{1}{\sqrt{10}} - i \frac{3}{\sqrt{10}}, -\frac{1}{\sqrt{10}} + i \frac{3}{\sqrt{10}}, -\frac{1}{\sqrt{10}} - i \frac{3}{\sqrt{10}}, \frac{3}{\sqrt{10}} + i \frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}} - i \frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}} + i \frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}} - i \frac{1}{\sqrt{10}} \}$$

However, only four of these turn out to also solve equation (2):

$$c + is \in \{ \frac{1}{\sqrt{10}}-i\frac{3}{\sqrt{10}}, -\frac{1}{\sqrt{10}}+i\frac{3}{\sqrt{10}}, \frac{3}{\sqrt{10}}+i\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}-i\frac{1}{\sqrt{10}} \}$$

So now we just need to multiply by $$r = \sqrt{5}$$ to get the roots.

$$z = r(c + is) \in \{ \frac{\sqrt{5}}{\sqrt{10}}-i\frac{3\sqrt{5}}{\sqrt{10}}, -\frac{\sqrt{5}}{\sqrt{10}}+i\frac{3\sqrt{5}}{\sqrt{10}}, \frac{3\sqrt{5}}{\sqrt{10}}+i\frac{\sqrt{5}}{\sqrt{10}}, -\frac{3\sqrt{5}}{\sqrt{10}}-i\frac{\sqrt{5}}{\sqrt{10}} \}$$ $$\boxed{z \in \{ \frac{1 - 3i}{\sqrt{2}}, \frac{-1+3i}{\sqrt{2}}, \frac{3+i}{\sqrt{2}}, \frac{-3-i}{\sqrt{2}} \}}$$

• You meant $c+is$ where you typed $c=is$, right? Commented May 3 at 21:56
• @J.W.Tanner: Yes. I just fixed it.
– Dan
Commented May 3 at 22:24