Question

find the root of complex number is 7+24i​

Answer 1

Answer:

hiii....

kaise ho.....

Answer 2

Step-by-step explanation:

Find the square root of 7-24i :

(a+bi)(a+bi)=7-24i

a^2-b^2+2abi=7-24i

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

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

(4-3i)(4-3i)=16-24i+9i^2=7-24i