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I got stuck while solving the following equation:$$z^6=(1+3i)^{12}.$$

I think that right side must (at least should) be transformed to the form$$z_{1}=1+3i=\sqrt{10}(\cos \theta+i\sin\theta ),$$and then we can use De Moivre's formula. However, I have no idea how to use it when $\cos\theta=\frac{1}{\sqrt{10}}$ and $\sin\theta=\frac{3}{\sqrt{10}}$. It is clear how to make power of complex numbers when $\theta$ is a "nice" angle, but what if it is not (and that is the case)? Thanks in advance!

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fdhd
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2 Answers2

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So, $\left(\frac z{(1+3i)^2}\right)^6=1$ or $\left(\frac z{-8+6i}\right)^6=1$ as $(1+3i)^2=1-9+6i=-8+6i$

Let $\frac z{-8+6i}=Y,z=Y(-8+6i)$

So, $Y^6=1$

This can be solved in 2 following ways

(i) $Y^6-1=(Y^3-1)(y^3+1)=(Y-1)(Y^2+Y+1)(Y+1)(Y^2-Y+1)$

If $Y-1=0,Y=1$

If $Y+1=0,Y=-1$

If $Y^2+Y+1=0, Y=\frac{-1\pm\sqrt{1-4}}2=\frac{-1\pm\sqrt3i}2$

If $Y^2-Y+1=0, Y=\frac{1\pm\sqrt{1-4}}2=\frac{1\pm\sqrt3i}2$

So, the six values of $Y$ are $\pm 1,\pm \frac{1-\sqrt3i}2, \pm \frac{1+\sqrt3i}2$

(ii)$Y^6=1=e^0=e^{2n\pi i}$ where $n$ is any integer.

So, $Y=e^{\frac{2n\pi i}6}=e^{\frac{n\pi i}3}$ where $n$ has $6$ in-congruent values $\pmod 6$ to provide $6$ distinct solutions.The simplest set of values of $n$ can be $0,1,2,3,4$ and $5$.

So, $Y=1,e^{\frac{\pi i}3}, e^{\frac{2\pi i}3}, e^{\frac{3\pi i}3}=e^{i\pi}=-1,e^{\frac{4\pi i}3}=e^{i\pi}\cdot e^{\frac{\pi i}3}=-e^{\frac{\pi i}3}$ and $e^{\frac{5\pi i}3}=-e^{\frac{2\pi i}3}$

So, the six values of $z$ are $\pm(-8+6i),\pm \left(\frac{1-\sqrt3i}2\right)(-8+6i), \pm \left(\frac{1+\sqrt3i}2\right)(-8+6i)$

So, $z^6=r^{12}e^{i 12\theta}$ where $r=\sqrt {10},$ and $\cos\theta=\frac 1{\sqrt{10}},\sin\theta=\frac 3{\sqrt{10}}$

So, $z=r^2e^{\frac{i 12\theta+2n\pi i}6}$ where $0\le n<6$

$z=10e^{2i\theta+\frac{n\pi i}3}=10(\cos(2\theta+\frac{n\pi}3)+i\sin(2\theta+\frac{n\pi}3))$

Now, $\cos2\theta=\cos^2\theta-\sin^2\theta=\frac{1-9}{10}=-\frac 4 5$ and $\sin2\theta=2\cos\theta\sin\theta=\frac 3 5$

Putting $n=0,z_0=10(\cos2\theta+i\sin2\theta)=10(-\frac45+i\frac 3 5)=-8+6i$

Putting $n=1,z_1=10(\cos(2\theta+\frac{\pi}3)+i\sin(2\theta+\frac{\pi}3))$

$\cos(2\theta+\frac{\pi}3)=\cos2\theta\cos\frac{\pi}3-\sin2\theta\sin\frac{\pi}3$ $=-\frac 4 5\cdot \frac 12-\frac 35\cdot \frac{\sqrt 3}2=-\frac{4+3\sqrt3}{10}$

$\sin(2\theta+\frac{\pi}3)=\cos2\theta\sin\frac{\pi}3+\sin2\theta\cos\frac{\pi}3$ $=-\frac 4 5\cdot\frac{\sqrt 3}2+\frac 35 \frac 12=\frac{3-4\sqrt3}{10}$

Putting $n=2,z_2$ $=10(\cos(2\theta+\frac{2\pi}3)+i\sin(2\theta+\frac{2\pi}3))$ $=-10(\cos(2\theta-\frac{\pi}3)+i\sin(2\theta-\frac{\pi}3))$ as $\cos(x+\frac{2\pi}3)=\cos(\pi+x-\frac{\pi}3)=-\cos(x-\frac{\pi}3)$ and so for sine.

$\cos(2\theta-\frac{\pi}3)=\cos2\theta\cos\frac{\pi}3+\sin2\theta\sin\frac{\pi}3$ $=-\frac 4 5\cdot \frac 12+\frac 35\cdot \frac{\sqrt 3}2=-\frac{4-3\sqrt3}{10}$

$\sin(2\theta-\frac{\pi}3)=\cos2\theta\sin\frac{\pi}3-\sin2\theta\cos\frac{\pi}3$ $=-\frac 4 5\cdot\frac{\sqrt 3}2-\frac 35 \frac 12=-\frac{3+4\sqrt3}{10}$

Now $10e^{2i\theta+\frac{(r+3)\pi i}3}=10e^{2i\theta+\frac{r\pi i}3}\cdot e^{i\pi}=-10e^{2i\theta+\frac{r\pi i}3}$ as $e^{i\pi}=-1$ i.e., $z_{r+3}=-z_r$

So, $z_5=-z_2,z_4=-z_1,z_3=-z_0$

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lab bhattacharjee
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A somewhat simpler approach: Certainly $z_0=(1+3i)^2$ is one solution. Now, in any equation $z^6=A$ for $A\ne0$, if one solution is $z_0$, the other solutions are $\zeta^iz_0$, where $\zeta$ is a primitive sixth root of unity and the exponents are in the range $1\le i\le5$. The natural choice for $\zeta$ is $\zeta=(1+\sqrt{-3})/2$. Writing down its powers is easy, and multiply these by $z_0=-8+6i$ to get all your solutions.

Lubin
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