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Question Number 20935 by Tinkutara last updated on 08/Sep/17
If ∣z + ω∣^2 = ∣z∣^2 + ∣ω∣^2 , where z and ω are complex numbers, then (1) (z/ω) is purely real (2) (z/ω) is purely imaginary (3) zω^� + z^� ω = 0 (4) amp((z/ω)) = (π/2)
$$\mathrm{If}\:\mid{z}\:+\:\omega\mid^{\mathrm{2}} \:=\:\mid{z}\mid^{\mathrm{2}} \:+\:\mid\omega\mid^{\mathrm{2}} ,\:\mathrm{where}\:{z}\:\mathrm{and}\:\omega \\ $$$$\mathrm{are}\:\mathrm{complex}\:\mathrm{numbers},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\frac{{z}}{\omega}\:\mathrm{is}\:\mathrm{purely}\:\mathrm{real} \\ $$$$\left(\mathrm{2}\right)\:\frac{{z}}{\omega}\:\mathrm{is}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\left(\mathrm{3}\right)\:{z}\bar {\omega}\:+\:\bar {{z}}\omega\:=\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{amp}\left(\frac{{z}}{\omega}\right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$
Answered by ajfour last updated on 08/Sep/17
∣z+ω∣^2 =∣z∣^2 +∣ω∣^2 ⇒ ∣(z/ω)+1∣^2 =∣(z/ω)∣^2 +1 but ∣(z/ω)+1∣^2 =∣(z/ω)∣^2 +1+2Re((z/ω)) so we can infer Re((z/ω))=0 ⇒ (z/ω) is purely imaginary. amp((z/ω)) = ±(π/2) Also (z+ω)(z^� +ω^� )=zz^� +ωω^� ⇒ zω^� +ωz^� =0 . (2) and (3), (4) are correct.
$$\mid{z}+\omega\mid^{\mathrm{2}} =\mid{z}\mid^{\mathrm{2}} +\mid\omega\mid^{\mathrm{2}} \\ $$$$\Rightarrow\:\mid\frac{{z}}{\omega}+\mathrm{1}\mid^{\mathrm{2}} \:=\mid\frac{{z}}{\omega}\mid^{\mathrm{2}} +\mathrm{1} \\ $$$${but}\:\mid\frac{{z}}{\omega}+\mathrm{1}\mid^{\mathrm{2}} =\mid\frac{{z}}{\omega}\mid^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{Re}\left(\frac{{z}}{\omega}\right) \\ $$$${so}\:{we}\:{can}\:{infer}\:\:\:{Re}\left(\frac{{z}}{\omega}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\frac{{z}}{\omega}\:{is}\:{purely}\:{imaginary}. \\ $$$${amp}\left(\frac{{z}}{\omega}\right)\:=\:\pm\frac{\pi}{\mathrm{2}} \\ $$$${Also}\:\left({z}+\omega\right)\left(\bar {{z}}+\bar {\omega}\right)={z}\bar {{z}}+\omega\bar {\omega} \\ $$$$\Rightarrow\:\:\:{z}\bar {\omega}+\omega\bar {{z}}\:=\mathrm{0}\:. \\ $$$$\left(\mathrm{2}\right)\:{and}\:\left(\mathrm{3}\right),\:\left(\mathrm{4}\right)\:\boldsymbol{{are}}\:\boldsymbol{{correct}}. \\ $$
Commented by Tinkutara last updated on 09/Sep/17
amp((z/ω)) = ±(π/2) so why (4) is correct?
$$\mathrm{amp}\left(\frac{{z}}{\omega}\right)\:=\:\pm\frac{\pi}{\mathrm{2}}\:\mathrm{so}\:\mathrm{why}\:\left(\mathrm{4}\right)\:\mathrm{is}\:\mathrm{correct}? \\ $$
Commented by Tinkutara last updated on 09/Sep/17
Thank you very much Sir! So this is one of the two solutions.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$$$\mathrm{So}\:\mathrm{this}\:\mathrm{is}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{solutions}. \\ $$
Commented by ajfour last updated on 09/Sep/17
no, you dont get what i mean amp((z/ω))=±(π/2), so amp((z/ω))=(π/2) is a possible solution.
$${no},\:{you}\:{dont}\:{get}\:{what}\:{i}\:{mean} \\ $$$${amp}\left(\frac{{z}}{\omega}\right)=\pm\frac{\pi}{\mathrm{2}},\:{so}\:{amp}\left(\frac{{z}}{\omega}\right)=\frac{\pi}{\mathrm{2}} \\ $$$${is}\:{a}\:{possible}\:{solution}. \\ $$

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