4th roots of complex number 7 + 24i

Question

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)

Actual Answer

Answer 1

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 $4 + 3 i$ $$ x^2 - y^2 = 4 \; , \; \; \; 2xy = 3 $$

Answer 2

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.

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