Solve the equation in z : 2z+6i = z/2i+5i-7.
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
We have to solve 2z + 6i = z/2i + 5i - 7
2z + 6i = z/2i + 5i - 7
=> 2z - z/ 2i = -i - 7
=> (2z*2i - z) / 2i = (-i - 7)
=> z( 4i - 1)/2i = (-i - 7)
=> z = (-i - 7) * 2i / ( 4i - 1)
=> z = (-2i^2 - 14i) / (4i - 1)
=> z = (2 - 14i)(4i + 1) / [(4i)^2 - (1)^2]
=> z = (8i + 56 + 2 - 14i) / (-16 - 1)
=> z = (58 - 6i)/ -17
=> z = -58/17 + 6i/17
The required result is z = -58/17 + 6i/17
giorgiana1976 | College Teacher | (Level 3) Valedictorian
We'll have to determine z = a + bi.
In other words, we'll have to determine the real part and the imaginary part of the complex number.
We'll move the terms in z to the left side:
2z - z/2i = 5i - 6i - 7
We'll combine like terms:
2z - z/2i = -i - 7
We'll multiply by 2i:
4iz - z = -2i^2 - 14i
z(4i - 1) = 2 - 14i
We'll divide by 4i - 1:
z = (2 - 14i)/(4i - 1)
We'll multiply by the conjugate of denominator:
z = (2 - 14i)(-1 - 4i)/(4i - 1)(-1 - 4i)
z = (-2 - 8i + 14i - 56)/(1+16)
z = (-58 + 6i)/17
z = -58/17 + (6/17)*i
The real part is: Re(z) = -58/17
The imaginary part is: Im(z) =6/17
z = -58/17 + (6/17)*i