Question

square root of 8-5i
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CLASS - 11
PLZZ ANSWER IT..... URGENT

Answer 1

Let $\bold{\sqrt{8-5i}=a+ib}$ where a is real part and b is imaginary part .
squaring both sides,
8 - 5i = a² + (ib)² + 2a.ib
⇒8 - 5i = a² - b² + i(2ab)
Compare both sides ,
a² - b² = 8 and 2ab = -5
(a² + b²)² = (a² - b²)² + 4a²b²
= 8² + (-5)² = 64 + 25 = 89
a² + b² = ±√89

Now, a² - b² + a² + b² = 8 ± √89
2a² = 8 ± √89 ⇒a² = 4 ± √(89/4)
a = √(4 + √(89/4)) = 2.95 find b=0.84
Then, a and b put in a + ib
a + ib is square root of 8 -5i
hence, 2.95+0.84i