Mathematics

Question

Find the multiplicative inverse of 6 + 2i

2 Answer

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    ________________


    Find the multiplicative inverse of

    [tex]\mathsf{z=6+2i}[/tex]

    ________


    The inverse multiplicative of  [tex]\mathsf{z=a+bi}[/tex]  is

    [tex]\mathsf{\dfrac{1}{z}}\\\\\\ =\mathsf{\dfrac{1}{a+bi}\qquad\quad(a\ne 0~~and~~b\ne 0)}\\\\\\ =\mathsf{\dfrac{1}{a+bi}\cdot \dfrac{a-bi}{a-bi}}\\\\\\ =\mathsf{\dfrac{1\cdot (a-bi)}{(a+bi)\cdot (a-bi)}}\\\\\\ =\mathsf{\dfrac{a-bi}{a^2-\,\diagup\hspace{-10}abi+\,\diagup\hspace{-10}abi-(bi)^2}}[/tex]

    [tex]=\mathsf{\dfrac{a-bi}{a^2-b^2\cdot i^2}}\\\\\\ =\mathsf{\dfrac{a-bi}{a^2-b^2\cdot (-1)}}\\\\\\ =\mathsf{\dfrac{a-bi}{a^2+b^2}}\\\\\\\\ \therefore~~\mathsf{\dfrac{1}{a+bi}=\dfrac{a}{a^2+b^2}-\dfrac{b}{a^2+b^2}\,i\qquad\quad\checkmark}[/tex]

    ________


    For this question,

    [tex]\mathsf{z=6+2i}[/tex]


    So,

    [tex]\mathsf{\dfrac{1}{z}}\\\\\\ =\mathsf{\dfrac{1}{6+2i}}\\\\\\ =\mathsf{\dfrac{6}{6^2+2^2}-\dfrac{2}{6^2+2^2}\,i}\\\\\\ =\mathsf{\dfrac{6}{36+4}-\dfrac{2}{36+4}\,i}\\\\\\ =\mathsf{\dfrac{6}{40}-\dfrac{2}{40}\,i}[/tex]


    [tex]\therefore~~\mathsf{\dfrac{1}{z}=\dfrac{3}{20}-\dfrac{1}{20}\,i}\quad\longleftarrow\quad\textsf{this is the answer.}[/tex]


    I hope this helps. =)

  • SOS

    Answer:

    [tex]\frac{3}{20}-\frac{1}{20}i[/tex]

    Step-by-step explanation:

    Find the multiplicative inverse of a complex number using the process described below:

    The inverse is found by reciprocating the original complex number. The reciprocal of the complex number (6+2i) is [tex]\frac{1}{6+2i}[/tex]. Multiply the numerator and denominator of the reciprocal by conjugate of the denominator and simplify:

    [tex]\frac{1}{6+2i}*\frac{6-2i}{6-2i}[/tex]

    You get: [tex]\frac{3}{20}-\frac{1}{20}i[/tex]

    Hope this helps!!

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