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Question-201629




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

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