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Answers
Given :-
The number 6-8i
To find :-
The square root of 6-8i
Solution :-
Given number = 6-8i
The square root of 6-8i = √(6-8i)
Let √(6-8i) = a + i b , Where a and b are real numbers
On squaring both sides then
=>[√(6-8i)]² = (a+ib)²
=> 6-8i = a²+2(a)(ib)+(ib)²
=> 6-8i = a²+i2ab+i²b²
=> 6-8i = a²+(2ab)i +(-b²)
Since, i² = -1
=> 6-8i = a²+(2ab)i -b²
=> 6-8i = (a²-b²) +(2ab)i
On comparing both sides then
a²-b² = 6 ----------(1)
and 2ab = -8
=> ab = -8/2
=> ab = -4
=> b = -4/a ---------(2)
On substituting the value of b in (1) then
=> a²-(-4/a)² = 6
=> a²-(16/a²) = 6
=> [(a²×a²)-16]/a² = 6
=> (a⁴-16)/a² = 6
=> a⁴-16 = 6a²
=> a⁴-6a²-16 = 0
=> a⁴-8a²+2a²-16 = 0
=> a²(a²-8)+2(a²-8) = 0
=> (a²-8)(a²+2) = 0
=> a²-8 = 0 or a²+2 = 0
=> a² = 8 or a² = -2
=> a² = 8 but a² ≠ 2
Since, a is a real number
=> a² = 8
=> a = ±√8
=> a = ±√(2×2×2)
=> a = ±2√2
Therefore, a = 2√2 and -2√2
If a = 2√2 then b = -4/(2√2)
=> b = -2/√2
=> b = (-√2×√2)/√2
=> b = -√2
If a = -2√2 then b = -4/(-2√2)
=> b = 2/√2
=> b = (√2×√2)/√2
=> b = √2
Therefore, a + ib = 2√2 -√2ior -2√2 +√2i
Therefore,
√(6-8i) = (2√2-√2 i) and -2√2 + √2 i
(or)
(2√2-√2 i) and -(2√2-√2 i)
Answer :-
The square root of 6-8i is (2√2-√2 i) and -(2√2-√2 i)
2√2+i√2. answer so this is the answer for you question
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