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

Step-by-step explanation:

We have , to simplify a complex expression . Basically a complex expression in the form a + ib , where i = $\sqrt{-1}$ 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 :

⇒ $Z = \sqrt{a^{2}+b^{2}}$

⇒ $Z = \sqrt{1^{2}+1^{2}}$

⇒ $Z = \sqrt{2}$

Putting value of $1+i$ as $Z = \sqrt{2}$ , we get:

⇒$2^{8}(1+i)^{8}$

⇒$2^{8}(\sqrt{2})^{8}$

⇒$256(2^{4})$

⇒$4096$