Question Number 164546 by mathls last updated on 18/Jan/22 | ||
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Commented by mathls last updated on 18/Jan/22 | ||
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$${Dom}\left({g}\right)=? \\ $$ | ||
Commented by mkam last updated on 18/Jan/22 | ||
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$$\boldsymbol{{D}}_{\boldsymbol{{g}}\left(\boldsymbol{{z}}\right)} \:=\:−\mathrm{2}\:<\:\boldsymbol{{z}}\:<\:\mathrm{2} \\ $$ | ||
Commented by mathls last updated on 19/Jan/22 | ||
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$${no}\:{find}\:{Dom}\left({g}\right) \\ $$ | ||
Commented by mkam last updated on 19/Jan/22 | ||
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$$\boldsymbol{{D}}_{\boldsymbol{{g}}\left(\boldsymbol{{z}}\right)} \:=\:\boldsymbol{{Dom}}\left(\boldsymbol{{g}}\right)\: \\ $$ | ||
Answered by mathmax by abdo last updated on 19/Jan/22 | ||
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$$\mid\mathrm{z}\mid<\mathrm{2}\:\:\mathrm{if}\:\mathrm{z}\:\mathrm{from}\:\mathrm{C}\:\:\:\mathrm{and}\:−\mathrm{2}<\mathrm{z}<\mathrm{2}\:\mathrm{if}\:\mathrm{z}\:\mathrm{is}\:\mathrm{real} \\ $$$$\mathrm{z}=\mathrm{x}+\mathrm{iy}\:\rightarrow\mid\mathrm{z}\mid<\mathrm{2}\:\Leftrightarrow\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }<\mathrm{2}\:\Rightarrow\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} <\mathrm{4}.... \\ $$ | ||