Math, asked by xayov50970, 8 months ago

If √ ( 5 + 12 i ) = a + ib , find (i). a² - b² , (ii). a² + b² (iii). find a and b (iv). √ ( 5 + 12 i )

Answers

Answered by praveenasg2007
0

Answer:

=(

−1

)

2

=−1

\sf{(a+ib)^{2}=a^{2}+(ib)^{2}+2abi=a^{2}-b^{2}+2abi}(a+ib)

2

=a

2

+(ib)

2

+2abi=a

2

−b

2

+2abi

Let \sf{\sqrt{5-12i}=x+iy}

5−12i

=x+iy

On squaring both the sides, we get

\sf{{(\sqrt{5-12i}})^{2} =(x+iy)^{2}}(

5−12i

)

2

=(x+iy)

2

\sf{5-12i=x^{2}-y^{2}+2xyi}5−12i=x

2

−y

2

+2xyi

On comparing both the sides, we get

Constant Term

\sf{x^{2}-y^{2}=5--------(1)}x

2

−y

2

=5−−−−−−−−(1)

Coefficient of i

\sf{2xy= -12}2xy=−12

\sf{y=\dfrac{-12}{2x}}y=

2x

−12

\sf{y=\dfrac{-6}{x}----------(2)}y=

x

−6

−−−−−−−−−−(2)

Now, on putting value of x in (1), we get

\sf{x^{2}-\bigg(\dfrac{-6}{x}\bigg)^{2}=5}x

2

−(

x

−6

)

2

=5

\sf{x^{2}-\bigg(\dfrac{36}{x^{2}}\bigg)=5}x

2

−(

x

2

36

)=5

\sf{\dfrac{x^{4}-36}{x^{2}}=5}

x

2

x

4

−36

=5

\sf{x^{4}-36=5x^{2}}x

4

−36=5x

2

\sf{x^{4}-5x^{2}-36=0}x

4

−5x

2

−36=0

\sf{x^{4}-9x^{2}+4x^{2} -36=0}x

4

−9x

2

+4x

2

−36=0

\sf{x^{2}(x^{2}-9)+4(x^{2} -9)=0}x

2

(x

2

−9)+4(x

2

−9)=0

\sf{(x^{2}+4)(x^{2}-9)=0}(x

2

+4)(x

2

−9)=0

\sf{x^{2}=9,-4}x

2

=9,−4

Since, x² can't be negative

So, x²≠ -4

x²= 9

x= ±3

When x=3, y= -2

x= -3, y= 2

So, required complex number is 3-2i and -3+2i.

Hence, square root of 5-12i is 3-2i or -3+2i.

I think useful thank you

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