Question

(1+i)^(-3) write in a+ib form.

Answer 1

Step-by-step explanation:

we Have Given That,

$( {1 + i})^{ - 3}$

which Is Equal To.,

$=> \frac{1}{ ({1 + i})^{3} }$

We Know,

$( {x + y})^{2} = {x}^{2} + {y}^{2} + 2xy$

Now,

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

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

Now,

We Know ,

${i}^{2} = - 1$

Put the Value of Iota square in Eqn..

we get,

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

$=> \frac{1}{2i + 2 {i}^{2} }$

$=> \frac{1}{2i - 2}$

Now Rationalising This Equation..

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

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

$=> \frac{2i + 2}{( - 4 - 4)}$

$=> \frac{2i + 2}{ - 8}$

Now separating These Two,

We Get (a+ib) Form, Which is Equal To....

Z=(a+ib)=

$- \frac{2i}{8} - \frac{2}{8}$

or

$- \frac{1}{4} - \frac{1}{4}i$

Hence, Solved...

Hope it Helped You!!!