# complex equation with a conjugate

Find all the solutions to the equation:
$$(1-i)z^{3}\bar{z}=7+i$$

My attempt:
$$z^3\bar{z}=\frac{7+i}{1-i}=3+4i\\$$
$$z^3\bar{z}=3+4i \rightarrow arg(z^3\bar{z})=arg(3+4i)\\$$
$$arg(z^3)-arg(z)=\tan^{-1}(4/3)+2\pi k\\$$
$$2arg(z)=\tan^{-1}(4/3)+2\pi k\\$$
Therefore the angle is:
$$arg(z)=\frac{\tan^{-1}(4/3)}{2}+\pi k$$

And to find the magnitude:
$$| z^3\bar{z}|=|z^3||z|=|z|^4=|3+4i| \\$$
$$|z|^4=5 \rightarrow |z|=\sqrt[4]5\\$$
Therefore
$$z=\sqrt[4]5\cdot e^{\frac{1}{2}i\tan^{-1}(4/3)+i\pi k}.$$
By plugging $$k = 0$$ and $$k = 1$$ we get
$$z_{1}=\sqrt[4]5\cdot e^{\frac{1}{2}i\tan^{-1}(4/3)}=\sqrt[4]5(\cos(0.5\tan^{-1}(4/3)) + i\sin(0.5\tan^{-1}(4/3))$$ $$z_{2}=\sqrt[4]5\cdot e^{\frac{1}{2}i\tan^{-1}(4/3)+i\pi}\\=\sqrt[4]5(\cos(0.5\tan^{-1}(4/3)+\pi) + i\sin(0.5\tan^{-1}(4/3)+\pi)$$

Yet the correct solution is:
$$z=\pm \sqrt[4]5(\cos(0.5\tan^{-1}(4/3)) + i\sin(0.5\tan^{-1}(4/3))\\$$
So $$z_1$$ is a correct solution, but i'm not sure what about $$z_2$$ and why there is a negative solution in the answer.
Thanks

Simplify $$z_2$$ using $$cos(\theta+\pi)=-cos(\theta)$$ and $$sin(\theta+\pi)=-sin(\theta)$$.

You can simplify $$z^3\bar z=\sqrt{5}z^2$$ giving $$z^2=\dfrac{3+4i}{\sqrt{5}}\iff z=\pm\dfrac{2+i}{\sqrt[4]{5}}$$

Since $$(2+i)^2=4+4i-1=3+4i$$

• Downvote??? This is an elegant solution, congrats @zwim (+1). Though, some additional steps can help to understand - would you add any? – user376343 Jan 20 at 23:36

From$$r^3r\text{ cis }(3\theta-\theta)=5\text{ cis}\arctan\frac43,$$ we draw $$r=\sqrt[4]5,$$

$$\theta=\frac12\arctan\frac43+k\pi.$$

Hence

$$\pm\sqrt[4]5\text{ cis}\left(\frac12\arctan\frac43\right).$$

(The $$\pm$$ comes from $$k\pi$$.)