Question

Compute each of the following, simplifying the result into a +bi form. I look at the answer in my book it is this. But How they solved it?

1) (2+2i)^8 =4096

Answer 1

Explanation:

We have , to simplify a complex expression . Basically a complex expression in the form a + ib , where i =
`√(-1)` and is called iota.

Expression:

`(2+2i)^(8)`

⇒
`(2+2i)^(8)`

⇒
`2^(8)(1+i)^(8)`

We know that the mod or modulus of a complex number in the form of
`a+ib` is represented by Z and Z =
`\sqrt{a^(2)+b^(2)}` . Computing mod of
`1+i` as :

⇒
`Z = \sqrt{a^(2)+b^(2)}`

⇒
`Z = \sqrt{1^(2)+1^(2)}`

⇒
`Z = √(2)`

Putting value of
`1+i` as
`Z = √(2)` , we get:

⇒
`2^(8)(1+i)^(8)`

⇒
`2^(8)(√(2))^(8)`

⇒
`256(2^(4))`

⇒
`4096`