What is the answer to (2z}{1+i}-\frac{2z)/i = 5/(2+i) ?

The answer to (2z}{1+i}-\frac{2z)/i = 5/(2+i) is z= 1/2-i 3/2

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(2z}{1+i}-\frac{2z)/i = 5/(2+i)

\frac{2 z}{1 + i} - \frac{2 z}{i} = \frac{5}{2 + i}

z = \frac{1}{2} - i \frac{3}{2}

Solution steps

\frac{2 z}{1 + i} - \frac{2 z}{i} = \frac{5}{2 + i}

Multiply by LCM

2 i (2 + i) z - 2 (1 + i) (2 + i) z = 5 i (1 + i)

Expand

2 i (2 + i) z - 2 (1 + i) (2 + i) z : - 2 i z - 4 z

Expand

5 i (1 + i) : - 5 + 5 i

- 2 i z - 4 z = - 5 + 5 i

Factor

- 2 i z - 4 z : - 2 (i + 2) z

- 2 (i + 2) z = - 5 + 5 i

Divide both sides by

- 2 (i + 2)

z = \frac{1}{2} - i \frac{3}{2}

\frac{\frac{3}{2}}{- \frac{1}{2}}

3 (\frac{16}{3})^{2} - 16 (\frac{16}{3})

\frac{x}{5} + \frac{8}{5} = 2

lo g_{7} (16)

x^{3} + 1 > 0

What is the answer to (2z}{1+i}-\frac{2z)/i = 5/(2+i) ?
The answer to (2z}{1+i}-\frac{2z)/i = 5/(2+i) is z= 1/2-i 3/2