Find the absolute values of z if 2z-i +5 = 3z -3i +1
Given the complex equation:
2z -i + 5 = 3z -3i +1
We need to find the absolute value of z.
First we need to rewrite z into the format of the complex number z = a+bi
let us combine terms with z on the left side.
==> 2z -3z = -3i +1 +i -5
==> -z = -4 -2i
Now we will multiply by -1
==> z = 4 + 2i
Now we will calculate the absolute values.
==> l z l = sqrt(a^2 + b^2)
= sqrt( 4^2 + 2^2)
= sqrt(16 + 4) = sqrt20 = 2sqrt5
==> l zl = 2sqrt5
We have to find the absolute value of z given that 2z-i +5 = 3z -3i +1
2z - i + 5 = 3z -3i +1
=> 3z - 2z = -i + 3i + 5 -1
=> z = 2i + 4
The absolute value of z is given by sqrt ( 2^2 + 4^2)
=> sqrt ( 4 + 16)
=> sqrt 20
Therefore the absolute value of z is sqrt 20.
We first solve the equation 2z-i +5 = 3z -3i +1 and then find the absolute value of z.
We subtract 3z from both sides:
2z-i+5-3z = = -3i+1.
-z -i+5 = -3i+1.
We add i-5 to both sides:
-z = -3i+1+i-5
-z = -2i-4.
We multiply both sides by -1.
z = 4+2i.
We know that the absolute vale of z = x+iy = sqrt(x^2+y^2).
Therefore the absolute value of z = (4+2i) is sqrt(4^2+2^2) = sqrt(16+4) = sqrt 20 = 2sqrt5.
Therefore the absolute value of z = 2sqrt5.