Question

Express the following complex numbers in the standard form a +ib:

solve step by step
solve 2nd and 3rd , I will mark you as brainlist​

Answer 1

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

Given complex number is

$\rm :\longmapsto\:\dfrac{3 + 2i}{ - 2 + i}$

So, on rationalizing the denominator, we get

$\rm \:  =  \: \dfrac{3 + 2i}{ - 2 + i} \times \dfrac{ - 2 - i}{ - 2 - i}$

$\rm \:  =  \: \dfrac{ - 6 - 3i - 4i - {2i}^{2} }{ {( - 2)}^{2} - {i}^{2} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {i}^{2} \: = \: - \: 1 \: }}}$

So, using this, we get

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

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

$\rm \:  =  \: \dfrac{ - 4 - 7i }{5}$

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

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

Given complex number is

$\rm :\longmapsto\:\dfrac{1}{ {(2 + i)}^{2} }$

$\rm \:  =  \: \dfrac{1}{4 + {i}^{2} + 2 \times 2 \times i}$

$\rm \:  =  \: \dfrac{1}{4 - 1 + 4i}$

$\rm \:  =  \: \dfrac{1}{3+ 4i}$

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

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

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

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {i}^{2} \: = \: - \: 1 \: }}}$

So, using this, we get

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

$\rm \:  =  \: \dfrac{3 - 4i}{9 + 16}$

$\rm \:  =  \: \dfrac{3 - 4i}{25}$

$\rm \:  =  \: \dfrac{3}{25} - \dfrac{4}{25}i$

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ADDITIONAL INFORMATION

$\purple{\rm :\longmapsto\:i \: = \: \sqrt{ - 1}}$

$\purple{\rm :\longmapsto\: {i}^{2} \: = \: - 1}$

$\purple{\rm :\longmapsto\: {i}^{3} \: = \: - i}$

$\purple{\rm :\longmapsto\: {i}^{4} \: = \: 1}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Complex \: number & \bf arg(z) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x + iy & \sf  {tan}^{ - 1}\bigg |\dfrac{y}{x} \bigg|   \\ \\ \sf  - x + iy & \sf \pi - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf  - x - iy & \sf  - \pi + {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf x - iy & \sf  - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \end{array}} \\ \end{gathered}$

Answer 2

Question:-

Express the following complex numbers in the standard form of a + ib:

$\sf \large ii) \frac{3 + 2i}{ - 2 + i} \\ \\ \sf \large iii) \frac{1}{(2 + i) {}^{2} }$

Solution:-

$\sf \large ii) \frac{3 + 2i}{ - 2 + i}$

Given:-

$\sf \large The \: complex \: number: \frac{3 + 2i}{ - 2 + i} .$

Multiplying and Dividing with – 2 – i.

$\sf \large⇒ a + ib = \frac{3 + 2i}{ - 2 + i} \times \frac{ - 2 - i}{ - 2 - i}$

$\sf \large⇒a + ib = \frac{3( - 2 - i) + 2i( - 2 - i)}{( - 2) {}^{2} - (i) {}^{2} }$

We know that, i ² = – 1.

$\sf \large⇒a + ib = \frac{ - 6 - 3i - 4i - 2i {}^{2} }{4 - i {}^{2} }$

$\sf \large⇒a + ib = \frac{ - 6 - 7i - 2( - 1)}{ 4 - ( - 1)}$

$\sf \large⇒a + ib = \frac{ - 4 - 7i}{5}$

$\sf \large { \boxed{ \sf \large \red{ \therefore The \: values \: of \: a,b \: are \: - \frac{4}{5} , - \frac{7i}{5} .}}}$

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Solution:-

$\sf \large iii) \frac{1}{(2 + i) {}^{2} }$

Given:-

$\sf \large The \: complex \: number: \frac{1}{ (2 + i) {}^{2} } .$

Now, let us simplify and Express in the standard form of ( a + ib ),

$\sf \large \frac{1}{2 + i) {}^{2} } = \frac{1}{(2 {}^{2} + i {}^{2} + 2(2)(i))}$

$\sf \large = \frac{1}{(4 - 1 + 4i)}$

We know that, i ² = – 1.

$\sf \large = \frac{1}{(3 + 4i)}$

[ By Multiplying and Dividing with (3 – 4i) ].

$\sf \large = \frac{1}{(3 + 4i)} \times \frac{(3 - 4i)}{(3 - 4i)}$

$\sf \large = \frac{(3 - 4i)}{(3 {}^{2} - (4i) {}^{2} )}$

$\sf \large = \frac{(3 - 4i)}{(9 - 16i {}^{2} )}$

$\sf \large = \frac{(3 - 4i)}{(9 - 16( - 1))}$

$\sf \large = \frac{(3 - 4i)}{25}$

Answer:-

$\sf \large { \boxed{ \sf \large \red{ \therefore The \: values \: of \: a,b \: are \: \frac{3}{25} , - \frac{4i}{25} .}}}$

Hope you have satisfied. ⚘