Question

Find square root of $7+24i$.

Answer 1

$\pm (4+3i)$

Answer 2

Let the square root of $7+24i$ be $x+iy$. Squaring both sides, we get: $7+24i = (x+iy)^2 = x^2 - y^2 + 2xyi$ Equating the real and imaginary parts:

1. $x^2 - y^2 = 7$
2. $2xy = 24 \implies xy = 12$

We also use the property $|x+iy|^2 = |7+24i|$. $x^2 + y^2 = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ Now we have a system of equations:

• $x^2 - y^2 = 7$
• $x^2 + y^2 = 25$

Adding these two equations: $2x^2 = 32 \implies x^2 = 16 \implies x = \pm 4$ Subtracting the first from the second: $2y^2 = 18 \implies y^2 = 9 \implies y = \pm 3$ From $xy=12$, we know that $x$ and $y$ must have the same sign. Thus, the possible pairs $(x,y)$ are $(4,3)$ and $(-4,-3)$. The square roots are $4+3i$ and $-4-3i$.