# what is the square root of the complex number 7-24i?

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...

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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`

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