Compute :
$(1 + i)^{- 1}$

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Solution

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

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Similar questions

(- 1)^{78}

11^{3}

1 = \sqrt{1} = \sqrt{(- 1) (- 1)} = \sqrt{- 1} \times \sqrt{- 1} = i \times i = - 1

A = [\begin{matrix} 2 & - 3 \\ - 4 & 7 \end{matrix}]

2 A^{- 1} = 9 I - A .

Now, can’t you compute these differences?

(i) 128 − 176

(ii) 248 − 312

(iii)

A^{2} - 4 A - 5

⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 2 2 & 1 & 1 2 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦

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