The square root of 3 - 4i is | vedclass

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(A) Let $\sqrt{3 - 4i} = x + iy$. Squaring both sides,we get $3 - 4i = (x^2 - y^2) + 2ixy$. Equating real and imaginary parts: $x^2 - y^2 = 3$ and $2x

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$\pm (2 - i)$

Vedclass · Answer

(A) Let $\sqrt{3 - 4i} = x + iy$. Squaring both sides,we get $3 - 4i = (x^2 - y^2) + 2ixy$. Equating real and imaginary parts: $x^2 - y^2 = 3$ and $2xy = -4$ (so $xy = -2$). We know that $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$. $(x^2 + y^2)^2 = (3)^2 + (-4)^2 = 9 + 16 = 25$. Thus,$x^2 + y^2 = 5$ (since $x^2 + y^2 > 0$). Adding $x^2 - y^2 = 3$ and $x^2 + y^2 = 5$,we get $2x^2 = 8$,so $x^2 = 4$,which means $x = \pm 2$. Subtracting the equations,we get $2y^2 = 2$,so $y^2 = 1$,which means $y = \pm 1$. Since $xy = -2$ (negative),$x$ and $y$ must have opposite signs. Therefore,the square roots are $\pm (2 - i)$.

The square root of $3 - 4i$ is

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