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3z - 2i = (5+i) / ( 1-i)
First we need to re-write z using the form z =a+ bi
First we will multiply and divide the right side by ( 1+ i)
==> 3z-2i = ( 5+ i)(1+i) / (1-i)(1+i)
==> 3z - 2i = ( 5 + 6i + i^2) / ( 1+ 1)
We know that i^2 = -1
==> 3z - 2i = 4 + 6i) / 2
==> 3z - 2i = 2 + 3i
Now let us add 2i to both sides:
==> 3z = 2+ 3i + 2i
==> 3z = 2 + 5i
Now divide by 3:
==> z = (2/3) + (5/3) i
Now we will calculate the absolute values:
l z l = sqrt( a^2 + b^2)
= sqrt( 2/3)^2 + ( 5/3)^2
= sqrt( 4+ 25)/9
- sqrt(29) / 3
==>The absolute value for z is : l Z l = sqrt29 / 3
To find the absolute value of z in 3x-2i = (5+i)/(1-i).
We have to solve this for z or 3z first in x+yi form and the find the absolute value of 3|z| = |x+yi| = sqrt(x^2+y^2) and |z| = (1/3)sqrt(x^2+y^2).
3z-2i = (5+i)/(1-i) = (5+i)(1+i)/(1-i)(1+i) = (5-1 +5i+i)/2 , as i^2= -1/
Therefore 3z -2i = (4+6i)/2
3z - 2i = 2+3i +2i = 2+5i.
Taking modulous (or absolute value) on both sides we get:
|3z| = 3|z| = |2+5i|
3|z|= sqrt(2^2+5^2) = sqrt29.
3|z| = sqrt29.
|z| = (sqrt29)/3.
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