How do you find $∣ ∣ \sqrt{5} + 2 i \sqrt{2} ∣ ∣$ $abs( sqrt 5 + 2i sqrt2)$ ?

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Answer 1 · 2016-04-29T13:51:46

$∣ ∣ \sqrt{5} + 2 \sqrt{2} i ∣ ∣ = \sqrt{13}$ $|sqrt5+2sqrt2i|=sqrt13$

$| a + b i | = \sqrt{a^{2} + b^{2}}$ $|a+bi|=sqrt(a^2+b^2)$

Hence, $∣ ∣ \sqrt{5} + 2 \sqrt{2} i ∣ ∣$ $|sqrt5+2sqrt2i|$

= $\sqrt{{(\sqrt{5})}^{2} + {(2 \sqrt{2})}^{2}}$ $sqrt((sqrt5)^2+(2sqrt2)^2)$

= $\sqrt{5 + 8}$ $sqrt(5+8)$

= $\sqrt{13}$ $sqrt13$

How do you find abs( sqrt 5 + 2i sqrt2)∣∣√5+2i√2∣∣abs( sqrt 5 + 2i sqrt2)?