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1 write each expression in the form $a + bi$, where $a$ and $b$ are rea…

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. $4 - 3\sqrt{-1}$

Explanation:

Response
Part a

Step1: Simplify the square root

Recall that \(\sqrt{-4}=\sqrt{4}\cdot\sqrt{-1} = 2i\) (since \(\sqrt{-1}=i\)).
So the expression becomes \(2\cdot2i\).

Step2: Multiply the coefficients

\(2\cdot2i = 4i\), which is in the form \(a + bi\) with \(a = 0\) and \(b=4\).

Step1: Multiply the coefficients and the imaginary units

First, multiply the coefficients \(3\) and \(2\), and multiply the \(i\) terms: \(3i\cdot2i=(3\times2)\cdot(i\times i)\).

Step2: Simplify \(i\times i\)

We know that \(i\times i=i^{2}=- 1\) (by the definition of the imaginary unit \(i\)). So \((3\times2)\cdot(i\times i)=6\cdot(-1)\).

Step3: Calculate the result

\(6\cdot(-1)=-6\), which is in the form \(a + bi\) with \(a=-6\) and \(b = 0\).

Step1: Recall the powers of \(i\)

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

Step2: Substitute \(i^{2}=-1\)

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

Step3: Calculate \((-1)^{2}\)

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

Answer:

\(0 + 4i\)

Part b