1

I came across this problem today and need some direction for part (b).

a) Show $\tan^{-1}\frac{24}{7} = 4\tan^{-1}\frac{1}{3}$.

b) Hence find the four fourth roots of $7 + 24i$ in Cartesian form.

Part a) is fine, since I used the fact that $Arg(z) = Arg(z_1z_2)$ on the R.H.S which equates to $Arg(z) = \tan^{-1}\frac{24}{7}$.

However, I'm unsure about Part b). I assume I use the fact that $Arg(z^n) = nArg(z)$.

My attempt is to find a root of the form (3 + i), but I'm unsure how $|3 + i| = \sqrt{10}$ finds its way into the answer.

Thanks in advance.

Marius S.L.
  • 2,245
Stephan
  • 469
  • 1
    Try to find a root of the form $z^4=c\cdot(3+i)$, and try to determine the $c>0$. Such that $|z|^4 = |7+24i|$. – DominikS Dec 15 '23 at 07:45
  • 1
    Thanks Dominik. You given me the direction I needed. I found $c$ to be $\frac{1}{\sqrt{2}}$ and then since we need 4 roots, multiply my first solution by $i$, ( a rotation of $\frac{\pi}{2}$ ) then multiply the second solution my $i$ and so on. Thanks – Stephan Dec 15 '23 at 21:49

0 Answers0