# Find the absolute value for z = (2-i) / -i

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### 2 Answers

z= (2-i)/ -i

First we need to rewrite the complext number z in the standard form which is z = a+ bi

Let us multiply and divide by the denominator inverse which is i

==> z = (2-i)*i / -i*i

= (2i - i^2) / -i^2

We know that:

i = sqrt-1 ==> i^2 = -1

==> z = (2i -(-1) / -(-1)

= (2i+1)/1

==> z = 2i + 1 = 1 + 2i

==> z = 1 + 2i

Now we will calculate the absolute values:

l z l = sqrt(a^2 + b^2)

= sqrt(1^2 + 2^2)

= sqrt(1+4) =sqrt5

**Then the absolute value for z = (2-i)/-i is : sqrt(5)**

To find the absolute value of z = (2-i)/i

Theabsolute value of a complex number x+yi = |x+yi| = sqrt(x^2+y^2).

Before findind |z = |(2-i)/i |, we bring z = (2-i)/i to the rectangular format x+yi.

We multipli both numerator and denominator of (2-i) by the conjugate of i, that is -i.

Then z = (2-i)i/(-i*i).

z = (2i-i^2)/(-i^2).

z = (2i+1)/1, as i^2 = -1.

Therefore z = 1+2i which is in x+yi format.

Therefore |z| = sqrt (1^2+2^2) = sqrt5.

Therefore the absolute value of (2-i)/i is |(2-i)/i| = sqrt5.