Question

multiplicative inverse of
root 5+2i /1- 2i​
plz send the answer fast
l want to submit it now

Answer 1

$\underline{\textsf{Given:}}$

$\dfrac{\sqrt{5}+2\,i}{1-2\,i}$

$\underline{\textsf{To find:}}$

$\textsf{Multiplicative inverse of \dfrac{\sqrt{5}+2\,i}{1-2\,i} }$

$\underline{\textsf{Solution:}}$

$\textsf{We know that,}$

$\textsf{The multiplicative inverse of z=a+b\,i is z^{-1}=\dfrac{1}{a+b\,i} }$

$\textsf{Let}\;z=\dfrac{\sqrt{5}+2\,i}{1-2\,i}$

$\textsf{The multiplicative inverse of z}$

$=\dfrac{1}{\frac{\sqrt{5}+2\,i}{1-2\,i}}$

$=\dfrac{1-2\,i}{\sqrt{5}+2\,i}$

$\textsf{To make it into standard form}$

$\textsf{multiply both numerator and denominator by \sqrt{5}-2\,ii }$

$=\dfrac{1-2\,i}{\sqrt{5}+2\,i}{\times}\dfrac{\sqrt{5}-2\,i}{\sqrt{5}-2\,i}$

$=\dfrac{(\sqrt{5}-2\,i)(1-2\,i)}{{\sqrt{5}}^2-2^2i^2}$

$=\dfrac{\sqrt{5}-2\sqrt{5}i-2i+4i^2}{5-4(-1)}$

$=\dfrac{\sqrt{5}-2\sqrt{5}i-2i+4(-1)}{9}$

$=\dfrac{\sqrt{5}-2\sqrt{5}i-2i-4}{9}$

$=\dfrac{(\sqrt{5}-4)+(-2\sqrt{5}-2)i}{9}$

$\underline{\textsf{Answer:}}$

$\textsf{The multiplicative inverse of \bf\dfrac{5+2\,i}{1-2\,i} is \bf\dfrac{(\sqrt{5}-4)+(-2\sqrt{5}-2)i}{9} }$

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Answer 2

Step-by-step explanation:

Given:

$z = \frac{ \sqrt{5} + 2i}{1 - 2i}$

To find: Multiplicative Inverse of z

Solution:

We know that multiplicative inverse of a/b is b/a

So,

multiplicative inverse of z is 1/z

$= \frac{1 - 2i}{ \sqrt{5} + 2i }$

Multiply and divide by complex conjugate of denominator

$= \frac{1 - 2i}{ \sqrt{5} + 2i} \times \frac{ \sqrt{5} - 2i }{ \sqrt{5} - 2i } \\ \\ = \frac{(1 - 2i)( \sqrt{5} - 2i)}{( \sqrt{5})^{2} + ( {2)}^{2} } \\ \\ = \frac{ \sqrt{5} - 2i - 2 \sqrt{5}i - 4 }{5 +4} \\ \\ \because \: {i}^{2} = - 1 \\ \\ = \frac{ \sqrt{5} - 4 - 2i (1 + \sqrt{5} )}{9} \\ \\ = \frac{ \sqrt{5} - 4}{9} -i\frac{ 2(1 + \sqrt{5} )}{9}$

Thus,

$\bold{\frac{ \sqrt{5} - 4}{9} -i\frac{ 2(1 + \sqrt{5} )}{9} }$

is the multiplicative inverse of Z.

Hope it helps you.

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