Question

Write z= 2+3i /1+2i in the form a +ib.(step by step)​

Answer 1

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

Given complex number is

$\rm :\longmapsto\:z = \dfrac{2 + 3i}{1 + 2i}$

On rationalizing the denominator, we get

$\rm :\longmapsto\:z = \dfrac{2 + 3i}{1 + 2i} \times \dfrac{1 - 2i}{1 - 2i}$

$\rm :\longmapsto\:z = \dfrac{2 + 3i - 4i - {6i}^{2} }{(1 + 2i)(1 - 2i)}$

We know,

$\boxed{ \tt{ \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

and

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

So, using this,

$\rm :\longmapsto\:z = \dfrac{2 - i - 6( - 1)}{ {1}^{2} - {(2i)}^{2} }$

$\rm :\longmapsto\:z = \dfrac{2 - i + 6}{ 1 - 4i^{2} }$

$\rm :\longmapsto\:z = \dfrac{8 - i }{ 1 + 4 }$

$\rm :\longmapsto\:z = \dfrac{8 - i }{5}$

$\rm :\longmapsto\:z = \dfrac{8}{5} \: - i \: \dfrac{1}{5}$

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Additional Information :-

$\boxed{ \tt{ \: i = \sqrt{ - 1} \: }}$

$\boxed{ \tt{ \: {i}^{3} = - i \: }}$

$\boxed{ \tt{ \: {i}^{4} = 1 \: }}$

$\boxed{ \tt{ \: \frac{1}{i} = - i \: }}$

$\boxed{ \tt{ \: {i}^{n} + {i}^{n + 1} + {i}^{n + 2} + {i}^{n + 3} = 0 \:if \: n \: is \: natural \: number }}$

Answer 2

Step-by-step explanation:

$complex number is</p><p>\rm :\longmapsto\:z = \dfrac{2 + 3i}{1 + 2i}:⟼z=1+2i2+3i</p><p>On rationalizing the denominator, we get</p><p>\rm :\longmapsto\:z = \dfrac{2 + 3i}{1 + 2i} \times \dfrac{1 - 2i}{1 - 2i}:⟼z=1+2i2+3i×1−2i1−2i</p><p>\rm :\longmapsto\:z = \dfrac{2 + 3i - 4i - {6i}^{2} }{(1 + 2i)(1 - 2i)}:⟼z=(1+2i)(1−2i)2+3i−4i−6i2</p><p>We know,</p><p>\boxed{ \tt{ \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}(x+y)(x−y)=x2−y2</p><p>and</p><p>\boxed{ \tt{ \: {i}^{2} \: = \: - \: 1 \: }}i2=−1</p><p>So, using this,</p><p>\rm :\longmapsto\:z = \dfrac{2 - i - 6( - 1)}{ {1}^{2} - {(2i)}^{2} }:⟼z=12−(2i)22−i−6(−1)</p><p>\rm :\longmapsto\:z = \dfrac{2 - i + 6}{ 1 - 4i^{2} }:⟼z=1−4i22−i+6</p><p>\rm :\longmapsto\:z = \dfrac{8 - i }{ 1 + 4 }:⟼z=1+48−i</p><p>\rm :\longmapsto\:z = \dfrac{8 - i }{5}:⟼z=58−i</p><p>\rm :\longmapsto\:z = \dfrac{8}{5} \: - i \: \dfrac{1}{5}:⟼z=58−i51</p><p>$