We know that \(\displaystyle i=\sqrt{-1}\) and \(\displaystyle i^2=-1\). \(\displaystyle \frac{i^5}{4}*\frac{16}{i^7}\) simplifies to \(\displaystyle \frac{4}{i^2}\). Since \(\displaystyle i^2=-1\) our final answer is \(\displaystyle \frac{4}{-1}=-4\).

To get rid of a complex number, we multiply by the conjugate which is opposite the sign.

\(\displaystyle \frac{1}{4+i}=\frac{1}{4+i}*\frac{4-i}{4-i}=\frac{4-i}{16-i^2}=\frac{4-i}{16-(-1)}=\frac{4-i}{17}\)

To get rid of a complex number, we multiply by the conjugate which is opposite the sign.

\(\displaystyle \frac{2-i}{3+i}=\frac{2-i}{3+i}*\frac{3-i}{3-i}=\frac{6+2i-3i-i^2}{9-i^2}=\frac{6-i-(-1)}{9-(-1)}=\frac{7-i}{10}\)

Write out some of the imaginary terms.

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

\(\displaystyle i^2= i\cdot i = -1\)

\(\displaystyle i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1\)

The powers of the imaginary number can be rewritten using the product of exponents.

\(\displaystyle i^{150} = (i^2)^{75} = (-1)^{75} = -1\)

\(\displaystyle i^8= (i^4)^2 = (1)^2 =1\)

\(\displaystyle i^{100} = (i^4)^{25} = (1)^{25} = 1\)

Replace all the terms back into the expression.

\(\displaystyle \frac{i^{150}-2i^{8}}{3i^{100}-6} = \frac{-1-2(1)}{3(1)-6} = \frac{-3}{-3} = 1\)

The answer is:  \(\displaystyle 1\)

The imaginary term \(\displaystyle i\) is equivalent to \(\displaystyle \sqrt{-1}\).

This means that:

\(\displaystyle i^6 =( i^2)^3 = (-1)^3 = -1\)

Substitute this term back into the numerator.

\(\displaystyle \frac{i^6+1}{i-1}=\frac{-1+1}{i-1} = \frac{0}{i-1} =0\)

There is no need to use extra steps such as multiplying by the conjugate of the denominator to simplify.

The answer is:  \(\displaystyle 0\)

Evaluate by first evaluating the imaginary terms.

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

\(\displaystyle i^2 = i\cdot i = -1\)

\(\displaystyle i^4= i^2\cdot i^2 =1\)

We can also use the product rule of exponents to simplify the higher powered imaginary numbers.

\(\displaystyle i^{12} = (i^4)^3=(1)^3 =1\)

\(\displaystyle \frac{2(2i^2-3)}{i^4-i^{12}}=\frac{2(2(-1)-3)}{1-1} =\frac{ -10}{0}\)

Note that evaluating the denominator will give us a zero denominator.  

This will indicate that the expression will approach to infinity.

The answer is:   \(\displaystyle \textup{The answer is not given.}\)

Write the first few terms of the imaginary term.

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

\(\displaystyle i^2 = i\cdot i =-1\)

\(\displaystyle i^3 = i^2\cdot i = -i\)

\(\displaystyle i^4 = i^2\cdot i^2=1\)

\(\displaystyle i^5 = (i^4)(i)=i\)

Notice that these terms will be in a pattern for higher order imaginary terms.

Rewrite the numerator using the product of exponents.

\(\displaystyle \frac{i^{60}-1}{i^3-1} =\frac{(i^4)^{15}-1}{i^3-1} = \frac{1-1}{-i-1} = 0\)

The answer is:  \(\displaystyle 0\)

Using the exponential property, we can expand the term in parentheses by multiplying the inner and outer powers together.

\(\displaystyle (7i^{-400})^{3} = (7^3\cdot i^{-400\times 3}) = 7^3 \cdot i^{-1200} = 343 \cdot i^{-1200}\)

Evaluate \(\displaystyle i^{-1200}\) by breaking this up as a product of imaginary powers.

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

\(\displaystyle i^2 = i\cdot i =-1\)

\(\displaystyle i^4 =i^2\cdot i^2 =1\)

\(\displaystyle i^{-1200} = (i^4)^{-300} = (1)^{-300} = 1\)

This means that:

\(\displaystyle 343 \cdot i^{-1200}= 343 \cdot 1\)

The answer is:  \(\displaystyle 343\)

Multiply the top and bottom by the conjugate of the denominator.

\(\displaystyle \frac{5i}{6-2i}\cdot \frac{6+2i}{6+2i}\)

Simplify the bottom using the FOIL method.

\(\displaystyle (6)(6)+(6)(2i)+(-2i)(6)+(-2i)(2i)\)

Simplify the expression by distributing all terms.  Recall that \(\displaystyle i=\sqrt{-1}\) and \(\displaystyle i^2=-1\).  Replace the term.

\(\displaystyle 36-4i^2 = 36-4(-1) = 36+4 =40\)

Simplify the top.

\(\displaystyle 5i(6+2i)=30i+10i^2 = 30i+(10)(-1) = 30i-10\)

Divide this by forty.

\(\displaystyle \frac{30i-10}{40} = \frac{3}{4}i-\frac{1}{4}\)

The answer is:  \(\displaystyle \frac{3}{4}i-\frac{1}{4}\)

Multiply the top and bottom by the conjugate of the denominator.

\(\displaystyle \frac{2i}{5-2i}\cdot \frac{5+2i}{5+2i}\)

Distribute the numerator and FOIL the denominator.

\(\displaystyle \frac{2i}{5-2i}\cdot \frac{5+2i}{5+2i} = \frac{10i+4i^2}{25-4i^2}\)

Recall that \(\displaystyle i=\sqrt{-1}\).  This means that \(\displaystyle i^2 = i\cdot i = -1\).

Replace the terms.

\(\displaystyle \frac{10i+4(-1)}{25-4(-1)} = \frac{10i-4}{29}\)

The answer is:  \(\displaystyle \frac{10i-4}{29}\)