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