Question

find the square root of 3-4i
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Answer 1

Answer:

you're d"mb so shut up bestie

Answer 2

Answer:

±(2+i)

Step-by-step explanation:

$Let \: \sqrt{3 - 4i} \: = \sqrt{x + i \sqrt{y} }$

$Then \: ({ \sqrt{3 + 4i} })^{2} = ( { \sqrt{x + i \sqrt{y} } })^{2}$

$⇒ \: 3 + 4i = x - y + 2i \sqrt{xy}$

$Comparing \: \: real \: \: part \: \: and \: \: imaginary \: \: part \: , \: \: we \: \: get$

$3 = x - y \: \: and \: \: 2 \sqrt{xy } = 4$

$⇒xy = 4$

$∴ \: x + y = ±5$

$As \: x≠ - 5 ; \: \: \sqrt{x} \: \: is \: \: a \: \: real \: \: part$

$∴ \: x = 4 \: \: and \: \: y = 1$

$⇒ \: \sqrt{x} = ±2 \: \: \: y =±1$

$Hence \: \: square \: \: root \: \: is \: \: ±(2 + i)$

Please mark me brainliest.