How could I find 79 as the sqrt of 6241 w/o using a calculator?

Question

Same with 3 - i being the sqrt of 8-6i

and

√2/2 (1+i) is the sqrt of i

Thanks!

Answer 1

1)

6241 = 6400 -160 + 1
6241 = 80^2 - 2(80) + 1
6241 = (80 -1)^2
6241 = 79^2

2)

(a + bi)^2 = a^2 - b^2 + 2abi

sqrt(8 - 6i)
8 is the difference of 2 squares, so factor 8.

8 = 1 * 8
(8 + 1)/2 = 9/2
(8 - 1)/2 = 7/2
8 = (9/2)^2 - (7/2)^2

8=2*4
(4+2)/2 = 3
(4 - 2)/2 = 1
8 = 3^2 - 1^2

-6 is twice the product of those 2 numbers.
-6=2(3*(-1))

a=3
b=-1

(3 - i)^2 = 8 - 6i

3)

The real part tells us

a^2 - b^2 = 0 so a = +/- b

The imaginary part tells us

2ab = 1

ab = 1/2

1/2 is positive so a=b and discard the negative possibility.

a^2 = 1/2
a = sqrt(2) / 2
b =sqrt(2) / 2

(sqrt(2) / 2 + i sqrt(2) / 2) ^2 = i

Note that these are the principle square roots, having real part greater than or equal to 0.

Answer 2

For square roots without a calculator see reference 1.

The square root of a complex number should be a complex number. So if you assume it is of the form (a + bi) then:
(a + bi)^2 = 8 - 6i
a^2 + 2abi - b^2 = 8 - 6i
so 2abi = -6i and ab = -3 or a = -3/b

Substitute "a" back into the previous equation:
9/b^2 - 6i - b^2 = 8 - 6i
9/b^2 - b^2 - 8 = 0
b^4 + 8b^2 - 9 = 0
(b^2 + 9)(b^2 - 1) = 0 and b^2 = 1 (using the real solution)
so b = -1 or b = 1 along with a=3 or a = -3
The square root is either: (3 - i) or (-3 + i)

Do as in the previous problem:
(a + bi)^2 = i
a^2 + 2abi - b^2 = i
2abi = i and a = 1/(2b)

Substitute "a" back into the previous equation:
1/(4b^2) - b^2 = 0
4b^4 = 1 and b = SQRT(1/2) or b = - SQRT(1/2)

This gives: a = 1/(2SQRT(1/2)) = SQRT(1/2) or a = - SQRT(1/2)
So the square roots of i are:
SQRT(1/2)(1 + i) or -SQRT(1/2)(1 + i)