Question

find the square root of 7 + 24 iota

Answer 1

Let the square root of 7 + 24i = a +bi where a and b are real numbers.

We know that i^2 = -1

Now (a+bi)^2 = 7 + 24i

=> a^2 -b^2 + 2(a)(b)i =7 + 24i

Comparing both sides we get two equations a^2 - b^2 =7

And 2ab = 24

=> ab = 12

=> b =12/a

Now putting the value of b in equation 1 we get

a^2 - (12/a)^2 = 7

=> a^2 -(144/a^2) = 7

=> a^4 -144 = 7a^2

=> a^4 - 7 a^2 -144 = 0

Solving above quadratic equation we get

a^2 = (7+√(576+49))/2 or a^2 = (7-√(576+49))/2

=> a^2 = (7+25)/2 or a^2= (7–25)/2

=> a^2 = 16 or a^2 = -9

Since we have assumed a as real a^2 can't be negative

Therefore a^2 = 16

=> a = 4 or a = -4

Putting the value of a in ab = 12 we get

b = 3 or b = -3

Therefore our required solution is 4 + 3i or -4 -3i