Question

find square root of complex number 3-4i.​

Answer 1

2=3−4

(+)2=3−4

2−2+2=3−4

Then got 2 simultaneous equations

2−2=3

and 2=−4

Solve for 2

in 1st equation: 2=3+2

Subbed into 2nd equation to power of 2

(3+2)2=4

4+32−4=0

Answer 2

Let 3 - 4i be the square root be a + bi, where a, b are real numbers. It says,

=> √(3 - 4i) = a + bi

=> 3 - 4i = (a + bi)^2

=> 3 - 4i = a^2 + (b)^2(i)^2 + 2abi

=> 3 - 4i = a^2 - b^2 + 2abi

Compare both sides(real & complex no)

=> 3 = a^2 - b^2 and - 4i = 2abi

=> 3 = a^2 - b^2 and (-2/b) = a

=> 3 = (-2/b)^2 - b^2

=> b^4 + 3b^2 - 4 = 0

=> (b^2 + 4)(b^2 - 1) = 0

Ignore b^2 + 4, as b is real number, b = ± 1

Therefore, a = (-2/±1) = ± 2.

Hence the square root of 3 - 4i is a + bi = 2 + 1i = 2 + i, ignoring the negative values of a and b.

Thus, √(3 - 4i) = 2 + i