Question

1. $(2i)^5 cdot (isqrt{6})^2$

Explanation:

Step1: Expand each power using exponent rules

For \((2i)^5\), we use \((ab)^n = a^n b^n\), so \((2i)^5 = 2^5 \cdot i^5 = 32 \cdot i^5\). For \((i\sqrt{6})^2\), we use \((ab)^n = a^n b^n\) and \((\sqrt{a})^2 = a\) (for \(a\geq0\)), so \((i\sqrt{6})^2 = i^2 \cdot (\sqrt{6})^2 = i^2 \cdot 6\).

Step2: Simplify \(i\) powers

Recall that \(i^2=-1\), \(i^3 = i^2 \cdot i=-i\), \(i^4=(i^2)^2 = 1\), and \(i^n\) cycles every 4. So \(i^5 = i^{4 + 1}=i^4 \cdot i = 1\cdot i = i\), and \(i^2=-1\).
Substitute these into our expressions: \((2i)^5 = 32i\) and \((i\sqrt{6})^2 = 6i^2=-6\).

Step3: Multiply the two simplified expressions

Now we multiply \(32i\) and \(-6\): \(32i\times(-6)=-192i\).

Answer:

\(-192i\)