Solving equation $(z-i)^3=(4-i\sqrt{48})z^3$ in $\mathbb C$.

Question

I need to solve the following equation in $\mathbb C$.

$$(z-i)^3=(4-i\sqrt{48})z^3.$$

I tried with trigonometric form , but having $z-i$ on the LHS is confusing me, since I get $(\cos\varphi+i\sin\varphi-i)^3$. On the RHS, I got $8\cdot\text{cis} (3\varphi+\frac{11\pi}6)$. I'm not sure how to handle the LHS, since nothing good seems to come from the naive $(a+b)^3$ approach.

score 5 · Accepted Answer · answered Oct 03 '13 at 20:34

5

Hint: Find cubic roots of $4-i\sqrt{48}=8e^{-i\pi/3}$.

answered Oct 03 '13 at 20:34

njguliyev

14,473
1
26
43

Oh, I see. Thanks. – Lazar Ljubenović Oct 03 '13 at 20:48

score 0 · Answer 2 · answered Oct 03 '13 at 20:41

0

Hint: try to find $r>0$ and $\phi\in[0,2\pi)$ such that $4-i\sqrt{48}=re^{i\phi}$. Then you get $\left(z-i\right)^{3}=\left(r^{1/3}e^{\left(1/3\right)i\phi}z\right)^{3}$ wich will bring you further.

answered Oct 03 '13 at 20:41

drhab

151,093

score 0 · Answer 3 · answered Oct 03 '13 at 21:46

0

Hint: solve $w^3=8\frac{1-i\sqrt{3}}2$ then solve $\frac{z-i}{z}=w$

answered Oct 03 '13 at 21:46

robjohn

345,667

Solving equation $(z-i)^3=(4-i\sqrt{48})z^3$ in $\mathbb C$.

3 Answers3