How do you find the absolute value of $4 - 8 i$ $4-8i$ ?

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Answer 1 · 2016-07-12T10:10:42

$| 4 - 8 i | = \sqrt{4^{2} + {(- 8)}^{2}} = \sqrt{16 + 64} = \sqrt{80} = 4 \sqrt{5} .$ $|4-8i|=sqrt{4^2+(-8)^2}=sqrt(16+64)=sqrt80=4sqrt5.$

Taking, $\sqrt{5} ≅ 2.236, ∣ z ⌈ = 4 \times 2.236 = 8.944$ $sqrt5~=2.236, |z|~=4xx2.236=8.944$

Absolute Value or Modulus $| z |$ $|z|$ of a Complex No. $z = x + i y$ $z=x+iy$ is defined by,

$| z | = \sqrt{x^{2} + y^{2}} .$ $|z|=sqrt(x^2+y^2).$

$| 4 - 8 i | = \sqrt{4^{2} + {(- 8)}^{2}} = \sqrt{16 + 64} = \sqrt{80} = 4 \sqrt{5}$ $|4-8i|=sqrt{4^2+(-8)^2}=sqrt(16+64)=sqrt80=4sqrt5$

Taking, $\sqrt{5} ≅ 2.236, ∣ z ⌈ = 4 \times 2.236 = 8.944$ $sqrt5~=2.236, |z|~=4xx2.236=8.944$

How do you find the absolute value of 4−8i4-8i?