Better Students Ask More Questions.
What is the absolute value of z if i(z-1)=-2 ?
3 Answers | add yours
To find the absolute value of the complex number, we'll put it in the rectangular form first.
For this reason, we'll re-write z, isolating z to the left side. For this reason, we'll remove the brackets:
iz - i = - 2
We'll add i both sides:
iz = i - 2
We'll divide by i both sides:
z = (i - 2)/i
Since we have to put z in the rectangular form and since we are not allowed to keep a complex number to the denominator, we'll multiply the ratio by the conjugate of i, that is -i.
z = -i*(i - 2)/-i^2
But i^2 = -1
z = -i*(i - 2)/-(-1)
We'll remove the brackets:
z = 2i - i^2
z = 1 + 2i
The modulus of z: |z| = sqrt (x^2 + y^2)
We'll identify x = 1 and y = 2.
|z| = sqrt(1 + 4)
The absolute value of the complex number z is: |z| = sqrt 5
Posted by giorgiana1976 on December 21, 2010 at 1:57 AM (Answer #1)
First let's determine z from i(z-1)=-2
=> iz - i = -2
=> iz = i - 2
=> z = i/i - 2/i
=> z = 1 - 2/i
=> z = 1 - 2i / i^2
=> z = 1 + 2i
Now the absolute value is sqrt ( 1^2 + 2^2) = sqrt (1 + 4) = sqrt 5.
The absolute value of z is sqrt 5.
Posted by justaguide on December 21, 2010 at 1:59 AM (Answer #2)
High School Teacher
To find the absolute value of z if i(z-1)=-2
We first solve for z and then determine the absolute value of z.
i(z-1) = -2.
z-1 = -2/i = -2*i/(i*i0 = -2i/(-1) = 2i.
Therefore z -1 = 2i
z = 2i+1 = 1+2i.
z = 1+2i.
Now absolute value of z= |x+yi| = (x^2+y^2)^(1/2).
|z| = |1+2i| = (1^2+2^2)^(1/2) = 5^(1/2).
So the absolute value of z = 5^(1/2).
Posted by neela on December 21, 2010 at 2:05 AM (Answer #3)
Join to answer this question
Join a community of thousands of dedicated teachers and students.