# Question a53f5

Feb 10, 2018

$Z = 1 + 0 i$

#### Explanation:

We have $\frac{1 + i}{1 - i} + \frac{2}{1 + i}$.

We must get a common denominator. This is $\left(1 - i\right) \left(1 + i\right)$.

$= \frac{\left(1 + i\right) \left(1 + i\right)}{\left(1 - i\right) \left(1 + i\right)} + \frac{2 \left(1 - i\right)}{\left(1 - i\right) \left(1 + i\right)}$

$= \frac{2 i}{2} + \frac{2 \left(1 - i\right)}{2}$

$= i + 1 - i$

$= 1$, or $1 + 0 i$

Feb 10, 2018

#### Explanation:

$Z = \frac{1 + i}{1 - i} + \frac{2}{1 + i}$

=>Z=((1+i)^2+2(1-i))/(1-i^2)" "["simple addition"]

$\implies Z = \frac{1 + 2 i + {i}^{2} + 2 - 2 i}{1 - {i}^{2}}$

$\implies Z = \frac{1 + 2 i - 1 + 2 - 2 i}{1 + 1} \text{ } \left[{i}^{2} = - 1\right]$

$\implies Z = 1$

=>color(red)(ul(bar(|color(green)(Z=1+0 cdot i)|#

Hope it helps...
Thank you..

Feb 10, 2018

$1$

#### Explanation:

You need to find a common denominator, add and simplify (understanding that ${i}^{2} = - 1$ )

$Z = \frac{1 + i}{1 - i} + \frac{2}{1 + i}$

$Z = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} + \frac{2}{1 + i} \times \frac{1 - i}{1 - i}$

$Z = {\left(1 + i\right)}^{2} / \left(\left(1 - i\right) \left(1 + i\right)\right) + \frac{2 \left(1 - i\right)}{\left(1 - i\right) \left(1 + i\right)}$

$Z = \frac{1 + 2 i + {i}^{2}}{1 - {i}^{2}} + \frac{2 - 2 i}{1 - {i}^{2}}$

$Z = \frac{1 + 2 i + {i}^{2} + 2 - 2 i}{1 - {i}^{2}}$

$Z = \frac{3 + {i}^{2}}{1 + 1}$

$Z = \frac{2}{2}$

$Z = 1$