Answer:
[tex]z_{3} =\frac{6}{5}+\frac{8}{5}i[/tex]
Step-by-step explanation:
As
[tex]z_{1} =2+i[/tex]
[tex]z_{2} =-2+4i[/tex]
[tex]\frac{1}{z_{3} }=\frac{1}{z_{1} }+\frac{1}{z_{2} }[/tex]
[tex]=\frac{1}{2+i}+\frac{1}{2\left(-1+2i\right)}[/tex]
So,
[tex]=\frac{2\left(-1+2i\right)}{2\left(2+i\right)\left(-1+2i\right)}+\frac{2+i}{2\left(2+i\right)\left(-1+2i\right)}[/tex]
[tex]\frac{1}{z_{3} }=\frac{2\left(-1+2i\right)+2+i}{2\left(2+i\right)\left(-1+2i\right)}[/tex]
As
[tex]2\left(2+i\right)\left(-1+2i\right)=-8+6i[/tex]
So,
[tex]\frac{1}{z_{3} }=\frac{2\left(-1+2i\right)+2+i}{-8+6i}[/tex]
[tex]\frac{1}{z_{3} }=\frac{5i}{-8+6i}[/tex]
[tex]\frac{1}{z_{3} }=\frac{3-4i}{10}[/tex]
Rewriting [tex]\frac{1}{z_{3} }=\frac{3-4i}{10}[/tex] in standard form
[tex]\frac{1}{z_{3} }=\frac{3}{10}-\frac{2}{5}i[/tex]
Substitute [tex]z_{3} =x+iy[/tex]
[tex]\frac{1}{x+iy }=\frac{3}{10}-\frac{2}{5}i[/tex]
As [tex]x+yi,\:10,\:5[/tex] has LCM [tex]10x+10iy[/tex]
[tex]\frac{1}{x+yi}\left(10x+10iy\right)=\frac{3}{10}\left(10x+10iy\right)-\frac{2}{5}i\left(10x+10iy\right)[/tex]
[tex]10=3\left(iy+x\right)-4i\left(iy+x\right)[/tex]
[tex]10=\left(3x+4y\right)+i\left(3y-4x\right)[/tex]
[tex]\mathrm{Rewrite\:}10\mathrm{\:in\:standard\:complex\:form:\:}10+0i[/tex]
[tex]10+0i=\left(3x+4y\right)+i\left(3y-4x\right)[/tex]
[tex]\mathrm{Complex\:numbers\:can\:be\:equal\:only\:if\:their\:real\:and\:imaginary\:parts\:are\:equal}[/tex]
[tex]\mathrm{Rewrite\:as\:system\:of\:equations:}[/tex]
[tex]\begin{bmatrix}10=3x+4y\\ 0=3y-4x\end{bmatrix}[/tex]
[tex]\begin{bmatrix}10=3x+4y\\ 0=3y-4x\end{bmatrix}:\quad y=\frac{8}{5},\:x=\frac{6}{5}[/tex]
[tex]\mathrm{Substitute\:back}\:z=x+yi[/tex]
[tex]z_{3} =\frac{6}{5}+\frac{8}{5}i[/tex]
Keywords: complex number
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