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