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Question-14244

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Question Number 14244 by Ruth1 last updated on 30/May/17
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/May/17
⊚10 ((jx)/(1+jy))=((3x+j4)/(x+3y))⇒x^2 j+3xyj=3x+j4+3xyj+4yj^2 ⇒(x^2 −4)−(3x−4y)=0⇒ { ((x^2 −4=0⇒x=±2)),((3x−4y=0⇒y=(3/4)(±2)=±(3/2) .■)) :} ⊚13 −1+j=(√2)(((−(√2))/2)+j((√2)/2))=(√2)(cos((7π)/4)+jsin((7π)/4))= =(√2).e^(j((7π)/4)) .■ ⊚6 z_1 =2+j,z_2 =−2+j4,(1/z_3 )=(1/z_2 )+(1/z_1 ) (1/z_1 )=(1/(2+j))×((2−j)/(2−j))=((2−j)/(4−j^2 ))=((2−j)/(4+1))=((2−j)/5) (1/z_2 )=(1/(−2+j4))×((−2−j4)/(−2−j4))=((−2−j4)/(4−j^2 16))=((−1−j2)/(10)) ⇒(1/z_3 )=((2−j)/5)+((−1−j2)/(10))=((4−j2−1−j2)/(10))= =((3−j4)/(10)) ⇒z_3 =((10)/(3−j4))×((3+j4)/(3+j4))=(2/5)(3+j4) .■ i cant read part #2 of this Q. ⊚11 z=((a+bj)/(c+dj))=((a+bj)/(c+dj))×((c−dj)/(c−dj))=((ac+bd−(ad−bc)j)/(c^2 +d^2 ))= =((ac+bd)/(c^2 +d^2 ))−((ad−bc)/(c^2 +d^2 ))j a)z in real if: ad−bc=0⇒ad=bc⇒(a/b)=(c/d) b)z is comlex when:ac+bd=0⇒(a/b)=−(d/c) .■
$$\circledcirc\mathrm{10} \\ $$$$\frac{{jx}}{\mathrm{1}+{jy}}=\frac{\mathrm{3}{x}+{j}\mathrm{4}}{{x}+\mathrm{3}{y}}\Rightarrow{x}^{\mathrm{2}} {j}+\mathrm{3}{xyj}=\mathrm{3}{x}+{j}\mathrm{4}+\mathrm{3}{xyj}+\mathrm{4}{yj}^{\mathrm{2}} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} −\mathrm{4}\right)−\left(\mathrm{3}{x}−\mathrm{4}{y}\right)=\mathrm{0}\Rightarrow \\ $$$$\begin{cases}{{x}^{\mathrm{2}} −\mathrm{4}=\mathrm{0}\Rightarrow{x}=\pm\mathrm{2}}\\{\mathrm{3}{x}−\mathrm{4}{y}=\mathrm{0}\Rightarrow{y}=\frac{\mathrm{3}}{\mathrm{4}}\left(\pm\mathrm{2}\right)=\pm\frac{\mathrm{3}}{\mathrm{2}}\:.\blacksquare}\end{cases} \\ $$$$\circledcirc\mathrm{13} \\ $$$$−\mathrm{1}+{j}=\sqrt{\mathrm{2}}\left(\frac{−\sqrt{\mathrm{2}}}{\mathrm{2}}+{j}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)=\sqrt{\mathrm{2}}\left({cos}\frac{\mathrm{7}\pi}{\mathrm{4}}+{jsin}\frac{\mathrm{7}\pi}{\mathrm{4}}\right)= \\ $$$$=\sqrt{\mathrm{2}}.{e}^{{j}\frac{\mathrm{7}\pi}{\mathrm{4}}} \:\:\:.\blacksquare \\ $$$$\circledcirc\mathrm{6} \\ $$$${z}_{\mathrm{1}} =\mathrm{2}+{j},{z}_{\mathrm{2}} =−\mathrm{2}+{j}\mathrm{4},\frac{\mathrm{1}}{{z}_{\mathrm{3}} }=\frac{\mathrm{1}}{{z}_{\mathrm{2}} }+\frac{\mathrm{1}}{{z}_{\mathrm{1}} } \\ $$$$\frac{\mathrm{1}}{{z}_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{2}+{j}}×\frac{\mathrm{2}−{j}}{\mathrm{2}−{j}}=\frac{\mathrm{2}−{j}}{\mathrm{4}−{j}^{\mathrm{2}} }=\frac{\mathrm{2}−{j}}{\mathrm{4}+\mathrm{1}}=\frac{\mathrm{2}−{j}}{\mathrm{5}} \\ $$$$\frac{\mathrm{1}}{{z}_{\mathrm{2}} }=\frac{\mathrm{1}}{−\mathrm{2}+{j}\mathrm{4}}×\frac{−\mathrm{2}−{j}\mathrm{4}}{−\mathrm{2}−{j}\mathrm{4}}=\frac{−\mathrm{2}−{j}\mathrm{4}}{\mathrm{4}−{j}^{\mathrm{2}} \mathrm{16}}=\frac{−\mathrm{1}−{j}\mathrm{2}}{\mathrm{10}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{z}_{\mathrm{3}} }=\frac{\mathrm{2}−{j}}{\mathrm{5}}+\frac{−\mathrm{1}−{j}\mathrm{2}}{\mathrm{10}}=\frac{\mathrm{4}−{j}\mathrm{2}−\mathrm{1}−{j}\mathrm{2}}{\mathrm{10}}= \\ $$$$=\frac{\mathrm{3}−{j}\mathrm{4}}{\mathrm{10}} \\ $$$$\Rightarrow{z}_{\mathrm{3}} =\frac{\mathrm{10}}{\mathrm{3}−{j}\mathrm{4}}×\frac{\mathrm{3}+{j}\mathrm{4}}{\mathrm{3}+{j}\mathrm{4}}=\frac{\mathrm{2}}{\mathrm{5}}\left(\mathrm{3}+{j}\mathrm{4}\right)\:\:.\blacksquare \\ $$$${i}\:{cant}\:{read}\:{part}\:#\mathrm{2}\:{of}\:{this}\:{Q}. \\ $$$$\circledcirc\mathrm{11} \\ $$$${z}=\frac{{a}+{bj}}{{c}+{dj}}=\frac{{a}+{bj}}{{c}+{dj}}×\frac{{c}−{dj}}{{c}−{dj}}=\frac{{ac}+{bd}−\left({ad}−{bc}\right){j}}{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }= \\ $$$$=\frac{{ac}+{bd}}{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }−\frac{{ad}−{bc}}{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }{j} \\ $$$$\left.{a}\right){z}\:{in}\:{real}\:{if}:\:{ad}−{bc}=\mathrm{0}\Rightarrow{ad}={bc}\Rightarrow\frac{{a}}{{b}}=\frac{{c}}{{d}} \\ $$$$\left.{b}\right){z}\:{is}\:{comlex}\:{when}:{ac}+{bd}=\mathrm{0}\Rightarrow\frac{{a}}{{b}}=−\frac{{d}}{{c}}\:.\blacksquare \\ $$

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