Question

drag each expression to the correct location on the table.
determine which expressions represent purely real numbers and which expressions represent non - real complex numbers.
$0 + 9\mathrm{i}$ $-12$ $\sqrt{(-5)^2}$ $2 - 7\mathrm{i}^2$ $7 - 5\mathrm{i}$ $\mathrm{i}^6$ $-\mathrm{i}^2 + \mathrm{i}^3$ $\sqrt{-6}$

purely real number	non - real complex number

Explanation:

Step1: Recall complex number form

A complex number is in the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. A purely real number has \(b = 0\), and a non - real complex number has \(b
eq0\) (or involves non - real components like square roots of negative numbers). Also, recall that \(i=\sqrt{- 1}\), \(i^{2}=-1\), \(i^{3}=i^{2}\cdot i=-i\), \(i^{4} = 1\), so \(i^{n}\) has a cycle of 4.

Step2: Analyze each expression

• For \(-12\):
• It can be written as \(-12 + 0i\), where \(a=-12\) and \(b = 0\). So it is a purely real number.
• For \(\sqrt{(-5)^{2}}\):
• First, calculate \((-5)^{2}=25\), then \(\sqrt{25}=5\), which can be written as \(5+0i\). So it is a purely real number.
• For \(2 - 7i^{2}\):
• Since \(i^{2}=-1\), substitute it in: \(2-7\times(-1)=2 + 7=9\), which is \(9+0i\). So it is a purely real number.
• For \(i^{6}\):
• Since \(i^{4}=1\), \(i^{6}=i^{4}\cdot i^{2}=1\times(-1)=-1\), which is \(-1 + 0i\). So it is a purely real number.
• For \(0 + 9i\):
• Here, \(a = 0\) and \(b=9

eq0\). So it is a non - real complex number.

• For \(7 - 5i\):
• Here, \(a = 7\) and \(b=-5

eq0\). So it is a non - real complex number.

• For \(-i^{2}+i^{3}\):
• First, \(i^{2}=-1\) and \(i^{3}=i^{2}\cdot i=-i\). Then \(-i^{2}+i^{3}=-(-1)+(-i)=1 - i\), where \(a = 1\) and \(b=-1

eq0\). So it is a non - real complex number.

• For \(\sqrt{-6}\):
• We can write \(\sqrt{-6}=\sqrt{6}\times\sqrt{-1}=\sqrt{6}i\), which is \(0+\sqrt{6}i\) with \(b=\sqrt{6}

eq0\). So it is a non - real complex number.

Answer:

Purely Real Number	Non - real Complex Number