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Question Number 71564 by 20190927 last updated on 17/Oct/19

$${(\mathrm{z}-\mathrm{i})}^4=-7+24\mathrm{i}$$

Commented by mathmax by abdo last updated on 18/Oct/19

$$wehave\mid -7+24i\mid =\sqrt{7^2+{24}^2}=\sqrt{49+576}=\sqrt{625}=25\Rightarrow$$$$-7+24i=25e^{-iarctan\left(\frac{24}{7}\right)}and{(z-i)}^4=25e^{-iarctan\left(\frac{24}{7}\right)}let$$$$z-i={re}^{i\theta }\Rightarrow r^4{e}^{i4\theta }=25e^{-iarctan\left(\frac{24}{7}\right)}\Rightarrow r{=}^4\sqrt{25}={\left(5^2\right)}^{\frac{1}{4}}=\sqrt{5}$$$$4\theta =-arctan\left(\frac{24}{7}\right)+2k\pi \Rightarrow {\theta }_k=-\frac{arctan\left(\frac{24}{7}\right)}{4}+\frac{k\pi }{2}$$$$andk\in \left[[0,3]\right]sotherootsofthisequationsare$$$$Z_k=\sqrt{5}{e}^{i(\frac{k\pi }{2}-\frac{arctan\left(\frac{24}{7}\right)}{4})}+iwith0⩽k⩽3$$$$$$$$$$

Commented by 20190927 last updated on 21/Oct/19

$$\mathrm{thank}\mathrm{you}$$

Answered by MJS last updated on 17/Oct/19

$$x^4=y\Rightarrow$$$$x_{1,2}=\pm \sqrt[4]{y}$$$$x_{3,4}=\pm \mathrm{i}\sqrt[4]{y}$$$$$$$$\sqrt[4]{-7+24\mathrm{i}}=2+\mathrm{i}$$$$\Rightarrow z-\mathrm{i}=\pm (2+\mathrm{i})\vee z-\mathrm{i}=\pm \mathrm{i}(2+\mathrm{i})$$$$\Rightarrow z=-2\vee z=2+2\mathrm{i}\vee -1+3\mathrm{i}\vee 1-\mathrm{i}$$

Commented by 20190927 last updated on 21/Oct/19

$$\mathrm{thank}\mathrm{you}$$$$$$

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