Question

1 write each expression in the form ( a + bi ), where ( a ) and ( b ) are real numbers. you may plot the numbers in the complex plane as a guide.
a. ( 2 cdot sqrt{-4} )
b. ( 3i cdot 2i )
c. ( i^4 )
d. ( 4 - 3sqrt{-1} )
2 which expression is equivalent to ( (3 + 9i) - (5 - 3i) )?
a ( -2 - 12i )
( -2 + 12i )
c ( 15 + 27i )
d ( 15 - 27i )

Explanation:

Problem 1 (Sub - parts a - d)

Part a: \(2\cdot\sqrt{- 4}\)

Step 1: Simplify the square root of a negative number

Recall that \(\sqrt{-n}=\sqrt{n}\cdot i\) for \(n>0\). So, \(\sqrt{-4}=\sqrt{4}\cdot i = 2i\).

Step 2: Multiply by 2

Now, \(2\cdot\sqrt{-4}=2\cdot(2i)=4i\), which is in the form \(a + bi\) with \(a = 0\) and \(b = 4\), so \(0+4i\).

Step 1: Multiply the coefficients and the imaginary units

First, multiply the coefficients \(3\times2 = 6\), and multiply the imaginary units \(i\times i=i^{2}\).

Step 2: Recall the value of \(i^{2}\)

We know that \(i^{2}=- 1\). So, \(3i\cdot2i=6\times i^{2}=6\times(-1)=-6\), which is in the form \(a + bi\) with \(a=-6\) and \(b = 0\), so \(-6+0i\).

Step 1: Recall the powers of \(i\)

We know that \(i^{1}=i\), \(i^{2}=-1\), \(i^{3}=i^{2}\cdot i=-i\), and \(i^{4}=(i^{2})^{2}\).

Step 2: Substitute the value of \(i^{2}\)

Since \(i^{2}=-1\), then \((i^{2})^{2}=(-1)^{2}=1\). So, \(i^{4}=1\), which is in the form \(a + bi\) with \(a = 1\) and \(b = 0\), so \(1+0i\).

Answer:

\(0 + 4i\)

Part b: \(3i\cdot2i\)