Square root of $3 - 4 i$ is

\pm (1 + 2 i)

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\pm (2 + i)

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\pm (1 - 2 i)

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\pm (2 - i)

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Solution

The correct option is D $\pm (2 - i)$
To find: $\sqrt{3 - 4 i}$

Solution:

$\sqrt{3 - 4 i} = \sqrt{(4 - 1) - 4 i}$

$\sqrt{3 - 4 i} = \sqrt{(2^{2} + i^{2}) - 4 i} . . . {i^{2} = - 1}$

$\sqrt{3 - 4 i} = \sqrt{2^{2} + i^{2} - 4 i}$

$\sqrt{3 - 4 i} = \sqrt{(2 - i^{2})}$

$\sqrt{3 - 4 i} = \pm (2 - i)$

Hence, square root of $3 - 4 i$ $= \pm (2 - i)$

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