Question Number 201629 by ali009 last updated on 09/Dec/23
Commented by ali009 last updated on 09/Dec/23
$${how}\:{is}\:{that}\:{cslculated}? \\ $$
Commented by aleks041103 last updated on 09/Dec/23
$${As}\:{far}\:{as}\:{I}\:{can}\:{tell}\:{this}\:{is}\:{an}\:{electrical} \\ $$$${engeneering}\:{problem}. \\ $$$${In}\:{EE}\:{they}\:{use}\:{j}\:{instead}\:{of}\:{i}\:{for}\:{the}\: \\ $$$${imaginary}\:{unit}\:\sqrt{−\mathrm{1}}. \\ $$$${Otherwise}\:{it}\:{is}\:{simple}\:{division}\:{of}\:{complex} \\ $$$${numbers}. \\ $$$${The}\:{most}\:'{difficult}'\:{part}\:{here}\:{is}\:{taking} \\ $$$${the}\:{squareroot}\:{of}\:{a}\:{complex}\:{number}. \\ $$$${Generally}\:{we}\:{use}\:{the}\:{polar}\:{form}\:{in}\:{such} \\ $$$${cases}. \\ $$$${z}={a}+{bj}={re}^{{j}\theta} \\ $$$${where}\:{r}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{and}\:\theta={arctan}\left({b}/{a}\right). \\ $$$${Then} \\ $$$$\sqrt{{a}+{bj}}=\sqrt{{re}^{{j}\theta} }=\sqrt{{r}}{e}^{{j}\theta/\mathrm{2}} \\ $$$${Now}\:{using}\:{Euler}'{s}\:{formula} \\ $$$$\sqrt{{a}+{bj}}=\sqrt{{r}}\left({cos}\left(\theta/\mathrm{2}\right)+{isin}\left(\theta/\mathrm{2}\right)\right)= \\ $$$$=\sqrt{{r}}{cos}\left(\theta/\mathrm{2}\right)+{i}\sqrt{{r}}{sin}\left(\theta/\mathrm{2}\right) \\ $$$${where}\:{again}\:{r}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{and}\:\theta={arctan}\left({b}/{a}\right). \\ $$$$ \\ $$$${The}\:{rest}\:{is}\:{just}\:{crunching}\:{the}\:{numbers}. \\ $$
Commented by ali009 last updated on 10/Dec/23
$${thank}\:{you}\:{so}\:{much}\:{and}\:{right}\:{this}\:{is} \\ $$$${communication}\:{engenniring}\:{problem} \\ $$$$ \\ $$