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Find-the-principal-value-of-3-4i-1-3-




Question Number 6923 by Tawakalitu. last updated on 03/Aug/16
Find the principal value of (3 + 4i)^(1/3)
$${Find}\:{the}\:{principal}\:{value}\:{of}\:\left(\mathrm{3}\:+\:\mathrm{4}{i}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$
Commented by Yozzii last updated on 03/Aug/16
3+4i=(√(3^2 +4^2 ))e^(itan^(−1) (4/3)) =5e^(itan^(−1) (4/3))   ∴ principal value of (3+4i)^(1/3)  is  (5e^(itan^(−1) (4/3)) )^(1/3) =5^(1/3) e^(i(1/3)tan^(−1) (4/3)) .  −−−−−−−−−−−−−−−−−−−−−−  z=re^(iθ) =re^(i(θ+2πk))    (k∈Z)  ⇒z^(1/n) =r^(1/n) e^(i((θ+2πk)/n))   (k=0,1,2,...,n−1)        principal value of z^(1/n)  occurs at k=0.
$$\mathrm{3}+\mathrm{4}{i}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }{e}^{{itan}^{−\mathrm{1}} \left(\mathrm{4}/\mathrm{3}\right)} =\mathrm{5}{e}^{{itan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}}} \\ $$$$\therefore\:{principal}\:{value}\:{of}\:\left(\mathrm{3}+\mathrm{4}{i}\right)^{\mathrm{1}/\mathrm{3}} \:{is} \\ $$$$\left(\mathrm{5}{e}^{{itan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}}} \right)^{\mathrm{1}/\mathrm{3}} =\mathrm{5}^{\mathrm{1}/\mathrm{3}} {e}^{{i}\frac{\mathrm{1}}{\mathrm{3}}{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}}} . \\ $$$$−−−−−−−−−−−−−−−−−−−−−− \\ $$$${z}={re}^{{i}\theta} ={re}^{{i}\left(\theta+\mathrm{2}\pi{k}\right)} \:\:\:\left({k}\in\mathbb{Z}\right) \\ $$$$\Rightarrow{z}^{\mathrm{1}/{n}} ={r}^{\mathrm{1}/{n}} {e}^{{i}\frac{\theta+\mathrm{2}\pi{k}}{{n}}} \:\:\left({k}=\mathrm{0},\mathrm{1},\mathrm{2},…,{n}−\mathrm{1}\right)\:\:\:\:\:\: \\ $$$${principal}\:{value}\:{of}\:{z}^{\mathrm{1}/{n}} \:{occurs}\:{at}\:{k}=\mathrm{0}. \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 03/Aug/16
I really appreciate
$${I}\:{really}\:{appreciate} \\ $$

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