Question

Find the multiplicative inverse of 6 + 2i

Answer 1

SOS

Answer:

$\frac{3}{20}-\frac{1}{20}i$

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 $\frac{1}{6+2i}$. Multiply the numerator and denominator of the reciprocal by conjugate of the denominator and simplify:

$\frac{1}{6+2i}*\frac{6-2i}{6-2i}$

You get: $\frac{3}{20}-\frac{1}{20}i$

Hope this helps!!

Answer 2

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________________


Find the multiplicative inverse of

$\mathsf{z=6+2i}$

________


The inverse multiplicative of  $\mathsf{z=a+bi}$  is

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

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

________


For this question,

$\mathsf{z=6+2i}$


So,

$\mathsf{\dfrac{1}{z}}\\\\\\ =\mathsf{\dfrac{1}{6+2i}}\\\\\\ =\mathsf{\dfrac{6}{6^2+2^2}-\dfrac{2}{6^2+2^2}\,i}\\\\\\ =\mathsf{\dfrac{6}{36+4}-\dfrac{2}{36+4}\,i}\\\\\\ =\mathsf{\dfrac{6}{40}-\dfrac{2}{40}\,i}$


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


I hope this helps. =)