"To determine: 1+i1−i+21+i?" solved and explained

Annette Sabin

Answered question

2022-01-16

To determine:

\frac{1 + i}{1 - i} + \frac{2}{1 + i}

Fasaniu

Beginner2022-01-17Added 46 answers

Step 1
We have

\frac{1 + i}{1 - i} + \frac{2}{1 + i}

We must get a common denominator. This is

(1 - i) (1 + i)

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

= \frac{2 i}{2} + ∣ \frac{2 (1 - i)}{2}

= i + 2 - i

= 1

1 + 0 i

MoxboasteBots5h

Beginner2022-01-18Added 35 answers

Step 1

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

\Rightarrow Z = \frac{{(1 + i)}^{2} + 2 (1 - i)}{1 - i^{2}}

simple addition

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

\Rightarrow Z = \frac{1 + 2 i - 1 + 2 - 2 i}{1 + 1} [i^{2} = - 1]

\Rightarrow Z = 1

\Rightarrow Z = 1 + 0 \times i

alenahelenash

Expert2022-01-24Added 556 answers

Step 1 You need to find a common denominator, add and simplify (understand 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 = \frac{(1 + i)^{2}}{(1 - i) (1 + i)} + \frac{2 (1 - i)}{(1 - i) (1 + i)}

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

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