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Question Number 127261 by mohammad17 last updated on 28/Dec/20
for any complex number if im(z)>0  then im((1/z))>0    prove this ?
$${for}\:{any}\:{complex}\:{number}\:{if}\:{im}\left({z}\right)>\mathrm{0} \\ $$$${then}\:{im}\left(\frac{\mathrm{1}}{{z}}\right)>\mathrm{0}\:\:\:\:{prove}\:{this}\:? \\ $$
Commented by mr W last updated on 28/Dec/20
wrong!  if im(z)>0 then im((1/z))<0.
$${wrong}! \\ $$$${if}\:{im}\left({z}\right)>\mathrm{0}\:{then}\:{im}\left(\frac{\mathrm{1}}{{z}}\right)<\mathrm{0}. \\ $$
Answered by mr W last updated on 28/Dec/20
z=a+bi  im(z)=b>0  (1/z)=(1/(a+bi))=((a−bi)/(a^2 +b^2 ))  im((1/z))=−(b/(a^2 +b^2 ))<0
$${z}={a}+{bi} \\ $$$${im}\left({z}\right)={b}>\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{{a}+{bi}}=\frac{{a}−{bi}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${im}\left(\frac{\mathrm{1}}{{z}}\right)=−\frac{{b}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }<\mathrm{0} \\ $$
Commented by mohammad17 last updated on 28/Dec/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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