How do you find absolute value of 4+2i, 8i, and -3 + 7i?

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How do you find absolute value of 4+2i, 8i, and -3 + 7i?

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Answer 1 · 2018-04-15T12:41:50

$2 \sqrt{5}, 8 and \sqrt{58}$ $2sqrt5,8" and "sqrt58$

Explanation:

$given a complex number x + y i$ $"given a complex number "x+yi$

$then the absolute value is$ $"then the absolute value is"$

$∙ x | x + y i | = \sqrt{x^{2} + y^{2}}$ $•color(white)(x)|x+yi|=sqrt(x^2+y^2)$

$4 + 2 i has x = 4 and y = 2$ $4+2i" has "x=4" and "y=2$

$\Rightarrow | 4 + 2 i | = \sqrt{4^{2} + 2^{2}} = \sqrt{20} = 2 \sqrt{5}$ $rArr|4+2i|=sqrt(4^2+2^2)=sqrt20=2sqrt5$

$8 i has x = 0 and y = 8$ $8i" has "x=0" and "y=8$

$\Rightarrow | 8 i | = \sqrt{0^{2} + 8^{2}} = 8$ $rArr|8i|=sqrt(0^2+8^2)=8$

$- 3 + 7 i has x = - 3 and y = 7$ $-3+7i" has "x=-3" and "y=7$

$\Rightarrow | - 3 + 7 i | = \sqrt{{(- 3)}^{2} + 7^{2}} = \sqrt{58}$ $rArr|-3+7i|=sqrt((-3)^2+7^2)=sqrt58$

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How do you find absolute value of 4+2i, 8i, and -3 + 7i?

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