This question tests one's ability to perform arithmetic operations on complex numbers. Questions like this introduces and builds on the concept of complex numbers. Recall that a complex number by definition contains a negative square. In mathematical terms this is expressed as follows.

\(\displaystyle \\i=\sqrt{-1} \\i^2=-1\)

Performing arithmetic operations on complex numbers relies on the understanding of the various algebraic operations and properties (distributive, associative, and commutative properties) as well as the imaginary, complex number \(\displaystyle i\).

For the purpose of Common Core Standards, "know there is a complex number \(\displaystyle i\) such that \(\displaystyle i^2=-1\), and every complex number has a form \(\displaystyle a+bi\) with \(\displaystyle a\) and \(\displaystyle b\) are reals", falls within the Cluster A of "perform arithmetic operations with complex numbers" (CCSS.MATH.CONTENT.HSF.CN.A).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Perform multiplication between the two terms.

\(\displaystyle \sqrt{-9}\times\sqrt{18}\)

Recall that multiplication between two radicand terms (terms under the square root sign), can be combined as one using the communicative property with multiplication.

\(\displaystyle \\ \sqrt{a}\times \sqrt{b}=\sqrt{a\times b} \\ \sqrt{-9}\times\sqrt{18}=\sqrt{-9\times 18}\)

Step 2: Factor the two term in the expression.

\(\displaystyle \\=\sqrt{-9\times 18} \\=\sqrt{-1\times 9\times 9\times 2}\)

Step 3: Pull out common terms that exists in the radicand.

Remember that when a number appears under the square root sign, one of the numbers can be brought out front and the other one is canceled out.

\(\displaystyle \sqrt{a\times a}= a\sqrt{1}=a\)

\(\displaystyle \\=\sqrt{-1\times 9\times 9\times 2} \\=9\sqrt{-1\times 2}\)

Step 4: Use the identity that \(\displaystyle \sqrt{-1}=i\).

\(\displaystyle \\=9\sqrt{-1\times 2} \\=9i\sqrt{2}\)

A pure imaginary number is expressed as \(\displaystyle bi\), where \(\displaystyle b\) is a positive real number and \(\displaystyle i\) represents the imaginary unit.

A pure imaginary number is expressed as \(\displaystyle bi\), where \(\displaystyle b\) is a positive real number and \(\displaystyle i\) represents the imaginary unit.

\(\displaystyle \sqrt{-225}=\sqrt{225}\cdot \sqrt{-1}\)

\(\displaystyle =15\cdot \sqrt{-1}\)

\(\displaystyle =15i\)

\(\displaystyle \sqrt{-98}=\sqrt{98}\cdot \sqrt{-1}\)

\(\displaystyle =\sqrt{49}\cdot \sqrt{2}\cdot \sqrt{-1}\)

\(\displaystyle =7\sqrt{2}\cdot \sqrt{-1}\)

\(\displaystyle =7\sqrt{2}\cdot i\)

\(\displaystyle =7i\sqrt{2}\)

\(\displaystyle \sqrt{-242}=\sqrt{242}\cdot \sqrt{-1}\)

\(\displaystyle =\sqrt{121}\cdot \sqrt{2}\cdot \sqrt{-1}\)

\(\displaystyle =11\cdot \sqrt{2}\cdot \sqrt{-1}\)

\(\displaystyle =11i\sqrt{2}\)

The powers of \(\displaystyle i\) are:

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

\(\displaystyle i^{2}=-1\)

\(\displaystyle i^{3}=-\sqrt{-1}\)

\(\displaystyle i^{4}=1\)

This pattern continues for every successive four power of \(\displaystyle i\). Thus:

\(\displaystyle i^{5}=\sqrt{-1}\)

\(\displaystyle i^{6}=-1\)

\(\displaystyle i^{7}=-\sqrt{-1}\)

\(\displaystyle i^{8}=1\)

The powers of \(\displaystyle i\) are:

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

\(\displaystyle i^{2}=-1\)

\(\displaystyle i^{3}=-\sqrt{-1}\)

\(\displaystyle i^{4}=1\)

This pattern continues for every successive four power of \(\displaystyle i\). Thus:

\(\displaystyle i^{5}=\sqrt{-1}\)

\(\displaystyle i^{6}=-1\)

\(\displaystyle i^{7}=-\sqrt{-1}\)

\(\displaystyle i^{8}=1\)

The powers of \(\displaystyle i\) are:

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

\(\displaystyle i^{2}=-1\)

\(\displaystyle i^{3}=-\sqrt{-1}\)

\(\displaystyle i^{4}=1\)

This pattern continues for every successive four power of \(\displaystyle i\). Thus:

\(\displaystyle i^{5}=\sqrt{-1}\)

\(\displaystyle i^{6}=-1\)

\(\displaystyle i^{7}=-\sqrt{-1}\)

\(\displaystyle i^{8}=1\)

The powers of \(\displaystyle i\) are:

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

\(\displaystyle i^{2}=-1\)

\(\displaystyle i^{3}=-\sqrt{-1}\)

\(\displaystyle i^{4}=1\)

This pattern continues for every successive four power of \(\displaystyle i\). Thus:

\(\displaystyle i^{5}=\sqrt{-1}\)

\(\displaystyle i^{6}=-1\)

\(\displaystyle i^{7}=-\sqrt{-1}\)

\(\displaystyle i^{8}=1\)

To simplify \(\displaystyle i\) to a larger power, simply break it into \(\displaystyle i^{4}\) terms, as these simplify to 1.

\(\displaystyle i^{50}=\left( i^{4}\right)^{12}\cdot i^{2}\)

\(\displaystyle =1\cdot i^{2}\)

\(\displaystyle =i^{2}\)

\(\displaystyle =-1\)