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Question Number 110608 by Engr_Jidda last updated on 29/Aug/20

$$\sqrt{x}+1=0$$$$Solution$$$$\sqrt{x}=-1$$$$recalledthat{\varrho }^{i\mathrm{\Pi }}=-1$$$$\sqrt{x}={\varrho }^{i\mathrm{\Pi }}$$$$x={\left({\varrho }^{i\mathrm{\Pi }}\right)}^2$$$$\Rightarrow x={\varrho }^{2i\mathrm{\Pi }}$$$$orx=2cos\mathrm{\Pi }+isin\mathrm{\Pi }$$$$xhasnorealvalue!$$

Commented by Her_Majesty last updated on 29/Aug/20

$${you}^{\prime }rewrong$$$$x=e^{2i\pi }\iff x=cos2\pi +isin2\pi =1$$$$anyway{you}^{\prime }rewrongbecausewearenot$$$$allowedtosquarebothsidesofanyequation;$$$$thisleadstofalsesolutions:$$$$x=3$$$$x^2=9\Rightarrow x=\pm 3$$$$$$$$inourcasewehavetosetrangesfirst$$$$ifx\in \mathbb{C}\Rightarrow x={re}^{i\theta }withr\in \mathbb{R}\wedge r⩾0and$$$$\theta \in \mathbb{R}\wedge \{\begin{array}{l}-\pi ⩽\theta <\pi \\ 0⩽\theta <2\pi \end{array}\left(bothareinuse\right)$$$$\sqrt{x}=-1$$$$\sqrt{{re}^{i\theta }}=-1$$$$\sqrt{r}{e}^{i\theta /2}=-1$$$$but-1=e^{\pm i\pi }$$$$\sqrt{r}{e}^{i\theta /2}=e^{\pm i\pi }\iff r=1\wedge \theta /2=\pm \pi \iff \theta =\pm 2\pi$$$$but\theta isnotinourrange\Rightarrow wehaveto$$$$subtract/add2\pi \Rightarrow \theta =0$$$$\Rightarrow x=\sqrt{x}=e^0=1\ne -1\Rightarrow nosolutionatall$$

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