How do you find $| 1 - 5 i |$ $abs(1 - 5i)$ ?

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How do you find $| 1 - 5 i |$ $abs(1 - 5i)$ ?

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Answer 1 · 2016-10-05T20:55:48

$\sqrt{26}$ $sqrt(26)$ .

Explanation:

Any complex number has the form $z = a + b i$ $z=a+bi$ , where $a$ $a$ is its real part and $b$ $b$ is its imaginary part. The norm of a complex number, $z$ $z$ , is defined as $| z | = \sqrt{a^{2} + b^{2}}$ $|z|=sqrt(a^2+b^2)$ .

In your case, the complex number $1 - 5 i$ $1-5i$ defines $a = 1$ $a=1$ and $b = - 5$ $b=-5$ . So, $a^{2} + b^{2} = 1 + 25 = 26$ $a^2+b^2 = 1+25=26$

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How do you find $| 1 - 5 i |$ $abs(1 - 5i)$ ?

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Humanities

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How do you find |1−5i|abs(1 - 5i)?

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How do you find $| 1 - 5 i |$ $abs(1 - 5i)$ ?