# Solve the equation in z : 2z+6i = z/2i+5i-7.

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### 2 Answers

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**

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**