Question

1.3+2i/-2+i
1/(2+i)²

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Answer 1

Answer:

Step-by-step explanation:

(i)

May be Your Question Is Incorrect

its 3+2i/-2+i

$\large\underline{\sf{Solution-}}$

Given complex number is

$\sf \longmapsto\: \bigg\lgroup\dfrac{ 3 + 2i}{ - 2 + i } \bigg \rgroup$

So, on rationalizing the denominator, we get

$\sf \: \longmapsto  =  \: \bigg \lgroup\dfrac{3 + 2i}{ - 2 + i} \times \dfrac{ - 2 - i}{ - 2 - i} \bigg \rgroup =$

$\sf \:  =  \longmapsto \bigg \lgroup \dfrac{ - 6 - 3i - 4i - {2i}^{2} }{ {( - 2)}^{2} - {i}^{2} } \bigg \rgroup$

We know,

$\begin{gathered} \rm \red\longmapsto\:\sf{ {i}^{2} \: = \: - \: 1 \: } \\ \end{gathered}$

So, using this, we get

$\sf \:  =  \: \dfrac{ - 6 - 7i - 2( - 1) }{ 4 - ( - 1)} = 4−(−1)−6−7i−2(−1)$

$\sf \:  =  \: \dfrac{ - 6 - 7i + 2}{ 4 + 1}$

$\sf \:  =  \: \dfrac{ - 4 - 7i }{5}$

$\sf \:  =  \: - \red{\dfrac{4}{5} - \dfrac{7}{5} i}$

(ii)

$\large\underline{\sf{Solution-}}$

Given complex number is

$\sf \longmapsto\:\dfrac{1}{ {(2 + i)}^{2} } \\ \longmapsto\sf \:  =  \: \dfrac{1}{4 + {i}^{2} + 2 \times 2 \times i}$

$\sf \:  = \longmapsto \: \dfrac{1}{4 - 1 + 4i}$

$\sf\:  =  \: \longmapsto\dfrac{1}{3+ 4i}$

$\sf \:  =  \:\longmapsto \dfrac{1}{3+ 4i}\times\dfrac{3 - 4i}{3 - 4i}$

$\sf \:  =  \:\longmapsto\dfrac{3 - 4i}{ {3}^{2} - {(4i)}^{2} }$

$\sf\longmapsto:  =  \: \dfrac{3 - 4i}{9 -16 {i}^{2} }$

We know,

$\begin{gathered} \sf \longmapsto\ \sf \boxed{ \sf {i}^{2} \: = \: - \: 1 \: } \\ \end{gathered}$

So, using this, we get

$\sf \:  =  \:\longmapsto \dfrac{3 - 4i}{9 -16( - 1)}$

$\sf \:  =  \:\longmapsto \dfrac{3 - 4i}{9 + 16} = 9+163−4i$

$\sf \:=\:\pink{\dfrac{3-4i}{25} }</p><p></p><p>$

$\sf \:  =  \: \boxed{ \red{\dfrac{3}{25} - \dfrac{4}{25}i }}$

❀тнαηк үσυ. . .!!❄︎

Answer 2

Answer:

Step-by-step explanation: