Question

Find square root of 7-24i

Answer 1

The real parts are equal, as are the real coefficients of the imaginary parts:

a^2-b^2=7

2abi=-24i

ab=-12

a=-12/b

 Substituting we get:

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

144/b^2-b^2=7

b^4+7b^2-144=0

(b^2+16)(b^2-9)=0

Since b is real b=3 or -3

a^2-9=7

a^2=16

a=+4,-4

Since ab=-12 one of a or b is negative.

So either of z=4-3i or z=-4 +3i  is a square root of 7-24i.

Answer 2

i.e; (cosx + i sinx)^n = cos nx + i sin nx 

[(-7)^2 + 24^2]^1/2 = 25 

-7+24i = 25 [-7/25 + i(24/25)] 

-7+24i = 25 [ cos x + i sin x], 

x = π - cos^-1 (7/25) = π-t (say) 

(-7+24i)^1/2 = 5 [ cos x + i sin x]^1/2 

= 5 [ sin t/2 + i cos t/2] 

cos t =7/25 = 2cos^2 (t/2) -1 

 cos(t/2) = 4/5 & hence sin (t/2) = 3/5 

(-7 + 24 i)^1/2 = 3 + 4 i