Question

if $i = \sqrt{-1}$, then $i^2 = \underline{\quad}$
even powers of $i = \underline{\quad}$, while odd powers of $i = \underline{\quad}$
a complex number is written in the form $\underline{\quad}$.

Explanation:

First blank (for \(i^2\)):

Step1: Recall the definition of \(i\)

Given \(i = \sqrt{-1}\), to find \(i^2\), we square both sides of the equation.

Step2: Calculate \(i^2\)

Squaring both sides, we get \(i^2=(\sqrt{-1})^2\). By the property of square roots, \((\sqrt{a})^2 = a\) for \(a\geq0\), here \(a = - 1\), so \(i^2=-1\).

Second part (Even and odd powers of \(i\)):

We know the cyclicity of \(i\):

• \(i^1=i\)
• \(i^2 = - 1\)
• \(i^3=i^2\times i=(-1)\times i=-i\)
• \(i^4=(i^2)^2=(-1)^2 = 1\)
• \(i^5=i^4\times i = 1\times i=i\) and so on.

For even powers: Let the power be \(2n\) where \(n\) is an integer. Then \(i^{2n}=(i^2)^n=(-1)^n\). When \(n\) is even, \((-1)^n = 1\); when \(n\) is odd, \((-1)^n=-1\). But in general, we can say that even powers of \(i\) are either \(1\) or \(- 1\) (depending on whether the power is divisible by \(4\) or not, but a more general description is that even powers of \(i\) are real numbers (either \(1\) or \(-1\)) and odd powers of \(i\) are either \(i\) or \(-i\) (imaginary numbers).

For odd powers: Let the power be \(2n + 1\) where \(n\) is an integer. Then \(i^{2n+1}=i^{2n}\times i=(-1)^n\times i\). So odd powers of \(i\) are imaginary numbers (either \(i\) or \(-i\)).

Third blank (Form of complex number):

A complex number is written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i=\sqrt{-1}\). Here, \(a\) is the real part and \(bi\) is the imaginary part.

Answer:

s:

• \(i^2=\boldsymbol{-1}\)
• Even powers of \(i\): real numbers (either \(1\) or \(-1\)), Odd powers of \(i\): imaginary numbers (either \(i\) or \(-i\))
• Complex number form: \(\boldsymbol{a + bi}\) (where \(a,b\in\mathbb{R}\))