Find all the values of $(14-3i)^{1/3}$.

Question

I have tried to find it, but I am not understanding what I should do. Please explain in detail. Thanks in advance.

Answer 1

Hint: if you cube $re^{i\theta}$ you get $r^3e^{3i\theta}$. You need to go the other direction. Remember you can wrap around by adding $2\pi$ to the angle.

Answer 2

You can think of the question as solving a complex polynomial equation: $$(14-3i)^{1/3} = ? \leftrightarrow 14-3i = z^3 \leftrightarrow z^3 - (14-3i) = 0$$ Since all complex polynomials with degree $n$ have $n$ complex roots (fundamental theory of algebra), the given polynomial has to have $3$ roots (maybe some with multiplicity $\ge 1$): $$\exists z_1,z_2,z_3 \in \Bbb{C}: \quad\begin{align} & z_1^3 - (14-3i) = 0 \\ & z_2^3 - (14-3i) = 0 \\ & z_3^3 - (14-3i) = 0 \end{align}$$ Here is the trick to find them: write up $z^3$ as: $$z^3 = (a+bi)^3 = (a^2+2abi-b^2)(a+bi) = a^3-3ab^2+3a^2bi-b^3i$$ Substitue $z^3 = 14-3i$ and group up the right side into real and imaginary terms: $$14-3i = (a^3-3ab^2)+(3a^2b-b^3)i$$ Since real and imaginary terms cannot mix, we get two equations: \begin{align}14 &= a^3-3ab^2 \\ 3 &= 3a^2b-b^3\end{align} Since as I've mentioned before, we are looking for $3$ roots, this will (unfortunately) lead to a cubic equation. (However, we know that all three solutions have to be real, which helps a lot.) So solving this, we get the answers of: $$a_1 \approx -1.35897, b_1 \approx 2.01236$$ $$a_2 \approx -1.06327, b_2 \approx -2.18308$$ $$a_3 \approx 2.42224, b_3 \approx 0.170721i$$ So our three complex solutions are: $$(14-3i)^{1/3} \approx \begin{align} & -1.35897 + 2.01236i, \\ & -1.06327 -2.18308i,& \\ & \quad 2.42224 + 0.170721i.\end{align}$$ Try raising any of these solutions to the $3$rd power! You'll get (approximately) $14-3i$.

Answer 3

we can let: $$z_0=14-3i$$ this can be represented in polar coordinates as follows: $$r=\sqrt{14^2+3^2}=\sqrt{205}$$ $$\theta=-\arctan\left(\frac{3}{14}\right)\approx-0.211$$ $$\therefore z_0=\sqrt{205}e^{-0.211i}$$ When we take roots of a complex number they will be equally spaced in a circle, with the same $r_i$ but a different $\theta_i$. These roots can be expressed as: $$\left(\sqrt{205}e^{-0.211i}\right)^{1/3}=205^{1/6}e^{\frac{-0.211}{3}+\frac{2\pi n}{3}}\,\,\,\,\,\,\,n=[0,1,2]$$ so we can say our three roots are: $$z_1=205^{1/6}\exp\left(-\frac{0.211}{3}\right)$$ $$z_2=205^{1/6}\exp\left(-\frac{0.211}{3}+\frac{2\pi}{3}\right)$$ $$z_3=205^{1/6}\exp\left(-\frac{0.211}{3}+\frac{4\pi}{3}\right)$$ Now you can work out the exact values if you need using: $$e^{ix}=\cos(x)+i\sin(x)$$