Math, asked by anasmohammadarif, 8 hours ago

Please help in finding the answer

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Answered by rajunaga110
1

Step-by-step explanation:

z^2 = (2^2-2*i*2+(I) ^2)

= 4-4i-1

=3-4i

w-i = 1+3i-i =1+2i

1/w = 1/(1+2i)

 =  \frac{1}{1 + 2i}  \times  \frac{1 - 2i}{1 - 2i}

 =  \frac{1 - 2i}{1 + 4}

 =  \frac{1 - 2i}{5}

so now

 \frac{ {z}^{2} }{w - i}  = (3 - 4i) \times (1 - 2i) \div 5

 =  \frac{(3 - 6i  - 4i + 8 {i}^{2} )}{5}

 =  \frac{3 - 8 - 10i}{5}  =   \frac{ -5 - 10i }{5}  =  - 1 - 2i

a+ib = -1-2i

conjugate of a+ib = -1+2i

modulus = √(1+4)=√5

Answered by mathdude500
4

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

Given complex numbers are

\rm :\longmapsto\:z = 2 - i

and

\rm :\longmapsto\:w = 1 + 3i

Now, Consider

\rm :\longmapsto\:\dfrac{ {z}^{2} }{w - i}

Let we assume that

\rm :\longmapsto\:p = \dfrac{ {z}^{2} }{w - i}

On substituting the values of z and w, we get

\rm :\longmapsto\:p = \dfrac{ {(2 - i)}^{2} }{1 + 3i - i}

\rm :\longmapsto\:p = \dfrac{ 4 + {i}^{2} - 4i }{1 + 2i}

\rm :\longmapsto\:p = \dfrac{ 4  - 1- 4i }{1 + 2i}

\red{\bigg \{ \because \:  {i}^{2}  \:  =  \:  -  \: 1\bigg \}}

\rm :\longmapsto\:p = \dfrac{ 3- 4i }{1 + 2i}

On rationalizing the denominator, we get

\rm :\longmapsto\:p = \dfrac{ 3- 4i }{1 + 2i} \times \dfrac{1 - 2i}{1 - 2i}

\rm :\longmapsto\:p = \dfrac{ 3 - 6i- 4i  +  {8i}^{2} }{ {1}^{2} -  {4i}^{2} }

\rm :\longmapsto\:p = \dfrac{ 3 - 10i   - 8}{ 1 + 4}

\rm :\longmapsto\:p = \dfrac{  - 5 - 10i}{5}

\bf\implies \:p =  - 1 - 2i

Hence,

\bf\implies \: \overline{p} =  - 1  +  2i

and

\bf\implies \: |p| =  \sqrt{ {( - 1)}^{2}  +  {( - 2)}^{2} } =  \sqrt{1 + 4} =  \sqrt{5}

Explore more :-

\boxed{ \tt{ \: z \: \overline{z} =  { |z| }^{2}  \: }}

\boxed{ \tt{ \:  |z|  \:  =  \:  |\overline{z}|  \: }}

 \boxed{ \tt{ \: |z_1 \: z_2|  =  |z_1|  |z_2|  \: }}

\boxed{ \tt{ \:  \bigg| \frac{z_1}{z_2}\bigg |  \:  =  \:  \frac{ |z_1| }{ |z_2| } \: }}

\boxed{ \tt{ \: \overline{z_1 + z_2} = \overline{z_1} + \overline{z_2} \: }}

\boxed{ \tt{ \: \overline{z_1  -  z_2} = \overline{z_1}  -  \overline{z_2} \: }}

\boxed{ \tt{ \: \overline{z_1   \times   z_2} = \overline{z_1}   \times   \overline{z_2} \: }}

\boxed{ \tt{ \: \overline{z_1  \div  z_2} = \overline{z_1}  \div  \overline{z_2} \: }}

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