Finding the complex number

Question

Find the complex number $z$ satisfying the system of equations $z^2 + \bar{w}^7=0$ and $z^5 w^{11}= 1$

Using Euler's form made it really complicated because we don't know the values of the modulus of z and w. Also, taking modulus on both sides of each equation didn't help too.

Answer is $i$

Note: $w$ is not cube root of unity.

Answer 1

Let $|z|=a$ and $|w|=b$.

We have $$z^5 w^{11}=1 \implies |z|^5 \cdot |w|^{11}=1 \implies \color{blue}{a^5b^{11}=1} \tag 1$$

Also,

$$z^2=-\bar{w}^{7} \implies |z|^2 = |w|^7 \implies \color{blue}{a^2=b^7} \tag 2$$

Using $(1)$ and $(2)$, we get $$|z|=|w|=1$$

Now, let $z=e^{i \theta} \quad ; \; w=e^{i \phi}$ and proceed.

Answer 2

Well, you have $z^5=w^{-11}$ and $z^2=-\bar{w}^7$.

Now $-2\times 2+5=1$, so $z=(z^2)^{-2}z^5=\bar{w}^{-14}w^{-11}.$

Answer 3

Say $z=r_1e^{i\theta_1}$ and $\omega=r_2e^{i\theta_2}$ Then from the second equation we have $$r_1^5r_2^{11}e^{i(5\theta_1+11\theta_2)}=1$$ This implies $$r_1^5r_2^{11}=1 \\ 5\theta_1+11\theta_2=2n_1\pi$$ Similarly we have $$z^2\bar\omega^{-7}=-1$$This implies $$r_1^2r_2^{-7}=1 \\ 2\theta_1-(-7\theta_2)=(2n_2+1)\pi$$ The values of $r_1=1$ and $r_2=1$ satisfy the equations in $r$'s. Solving simultaneously for $\theta$'s$$\theta_1=\frac{1}{13}(14n_1-22n_2-11)\pi \\ \theta_2=\frac1{13}(-4n_1+10n_2+5)\pi$$For $n_1,n_2\in\mathbb{N}$

Let's put $n_1=1$ and $n_2=1$ we have $\theta_1=\frac{-19\pi}{13}$ and $\theta_2=\frac{11\pi}{13}$

Put this in the first equation $$z^2\bar\omega^{-7}=\exp(\frac{-38\pi i}{13})\exp(\frac{77\pi i}{13})=\exp(3\pi i)=-1$$ and the second equation $$z^5\omega^{11}=\exp(\frac{-95\pi i}{13})\exp(\frac{121\pi i}{13})=\exp(2\pi i)=1$$