solve-for-z-C-z-a-bi-z-a-bi-r-R-r-2-z-2-z-r-2-z-2-z-

Question Number 76397 by MJS last updated on 27/Dec/19

$$\mathrm{solve}\:\mathrm{for}\:{z}\in\mathbb{C} \\ $$$$\left[{z}={a}+{b}\mathrm{i};\:\bar {{z}}={a}−{b}\mathrm{i};\:{r}\in\mathbb{R}\right] \\ $$$$\sqrt{{r}^{\mathrm{2}} −{z}^{\mathrm{2}} }=\bar {{z}} \\ $$$$\sqrt{{r}^{\mathrm{2}} +{z}^{\mathrm{2}} }=\bar {{z}} \\ $$

Commented by mr W last updated on 27/Dec/19

$${is}\:{there}\:{a}\:{solution}\:{other}\:{than} \\ $$$${z}=\mathrm{0}\:{and}\:{r}=\mathrm{0}? \\ $$

Commented by MJS last updated on 27/Dec/19

$$\mathrm{I}\:\mathrm{found}\:\mathrm{no}\:\mathrm{other}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{one} \\ $$$$\sqrt{{r}^{\mathrm{2}} −{z}^{\mathrm{2}} }=\bar {{z}}\:\Rightarrow\:{z}=\frac{\sqrt{\mathrm{2}\left({r}^{\mathrm{2}} +\mathrm{2}{s}^{\mathrm{2}} \right)}}{\mathrm{2}}+{s}\mathrm{i}\:\mathrm{with}\:{r},\:{s}\:\in\mathbb{R} \\ $$

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