Question

find the square root of the following complex number-48-14i​

Answer 1

Let $-48-14i$ be a square of a complex number $a+bi$.

$a$, $b$ ∈ R (R is a set of real numbers)

Square of $a+bi$ :  $(a+bi)^2=a^2-b^2+2abi$

$a^2-b^2+2abi=-48-14i$

Know that :

(Real part of LHS) = (Real part of RHS)

(Imaginary part of LHS) = (Imaginary part of RHS)

$a^2-b^2=-48$ --- 1

$2ab=-14$ --- 2

Know that $b=-\frac{7}{a}$ --- 3 ( ∵ 2 )

Substitute into 1.

$a^2-\frac{49}{a^2} =-48$

Know that a > 0, b < 0 or a < 0, b > 0 --- 4 ( ∵ 1, 2 )

Multiply a² to both sides.

$a^4+48a^2-49=0$

$(a^2-1)(a^2+49)=0$

∴ $a=\pm 1$ ( ∵ 4 )

∴ $b=\mp 7$ ( ∵ 3 )

The complex number is $\pm 1 \mp 7i$.

∴  $\sqrt {-48-14i}=1-7i$ or $-1+7i$

Sorry for suggesting wrong answer $7-i$.

I confused the sign of the coefficient of polynomial term.

It gave me wrong value of a, sorry!

T H A N K

Y O U !

Answer 2

answer is attached in the given attachment