Question bank — Imaginary unit i = √(−1), i² = −1, powers of i cycle

$i$ is a real number that we just haven't found on the number line yet. ::: False. Every real number squared is $\ge 0$, but $i^2=-1<0$. No real number can do this, so $i$ genuinely lives off the number line — it is not "hidden" somewhere on it. Because $i^2=-1$, the number $i$ is "less than zero." ::: False. "Less than" (ordering) only makes sense for real numbers. $i$ is not on the number line, so asking whether it is positive or negative has no meaning; only its square is negative. $i^4=1$, so $i$ behaves exactly like the real number $1$. ::: False. $i^4=1$ only says four multiplications by $i$ return you to the start. $i$ itself is still $i$, not $1$; $i^1=i\ne 1$. The four values $\{1,i,-1,-i\}$ always add to zero. ::: True. $1+i+(-1)+(-i)=0$: the reals cancel ($1-1=0$) and the imaginaries cancel ($i-i=0$). This is why any four consecutive powers of $i$ sum to $0$. $\sqrt{-1}\cdot \sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt{1}=1$. ::: False. The rule $\sqrt{a}\sqrt{b}=\sqrt{ab}$ breaks when both are negative. Correctly, $\sqrt{-1}\cdot \sqrt{-1}=i\cdot i=i^2=-1$. Multiplying a point by $i$ changes how far it is from the origin. ::: False. Multiplying by $i$ is a pure $90^{\circ }$ rotation; it changes the direction but not the distance from the origin (the modulus is unchanged). Every complex number $a+bi$ has both a real part and an imaginary part, so $7$ is not complex. ::: False. $7=7+0i$: its imaginary part is just $0$. Every real number is a complex number with zero imaginary part. $i^{-1}$ is undefined because you can't divide by an imaginary number. ::: False. $i^{-1}=\frac{1}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i$, a perfectly valid number. Division by $i$ is fine as long as you clear it from the denominator.

"$i^3=i\cdot i\cdot i=i$, because there are three $i$'s." ::: The error is not collapsing $i\cdot i=i^2=-1$. Correctly $i^3=(i^2)\cdot i=(-1)i=-i$, not $i$. "$i^0$ is meaningless because $i$ is imaginary." ::: Wrong: any nonzero base to the power $0$ is $1$, and $i\ne 0$, so $i^0=1$. The base being imaginary changes nothing about the zero-exponent rule. "$\sqrt{-9}=-3$ because a negative goes with a negative." ::: Wrong: $\sqrt{-9}=\sqrt{9}\cdot \sqrt{-1}=3i$. You must factor out $\sqrt{-1}=i$; the answer is imaginary, not the real number $-3$. "$i^{100}$ is huge because it is $i$ raised to a big power." ::: Wrong: powers of $i$ cycle through only four values, so $i^{100}=i^{100\mathrm{mod}4}=i^0=1$. Size of the exponent is irrelevant — only the remainder matters. "To find $i^{27}$, since $27$ is odd the answer must be $\pm i$." ::: Partly lucky, wrong reasoning. Oddness alone isn't enough: you need $27\mathrm{mod}4=3$, giving $i^3=-i$. (Odd exponents can also give $+i$, e.g. $i^5=i$.) "$\sqrt{-4}\cdot \sqrt{-9}=\sqrt{36}=6$." ::: The rule $\sqrt{a}\sqrt{b}=\sqrt{ab}$ fails for two negatives. Convert first: $(2i)(3i)=6i^2=-6$. "$-i$ is not one of the powers of $i$ — the cycle is $1,i,-1$." ::: Wrong: the cycle has four members $1,i,-1,-i$. $-i$ appears as $i^3$ (and $i^7$, etc.). "Since $i=\sqrt{-1}$, we also have $-i=\sqrt{-1}$, so $i=-i$." ::: Wrong: both $i$ and $-i$ square to $-1$, but they are different numbers. We choose $i$ to be the principal one; $i=-i$ would force $2i=0$, i.e. $i=0$, which is false.

Why is the period of the powers of $i$ exactly $4$, not $2$ or $8$? ::: Because $i^4=(i^2{)}^2=(-1{)}^2=1$ is the first power that returns to $1$, and geometrically four $90^{\circ }$ turns make a full $360^{\circ }$ circle. Why does only the remainder $n\mathrm{mod}4$ decide the value of $i^n$? ::: Write $n=4k+r$. Then $i^n=(i^4{)}^k\cdot i^r=1^k\cdot i^r=i^r$: the full loops of four contribute a factor of $1$ and vanish, leaving only $i^r$. Why must we convert $\sqrt{-x}$ to $i\sqrt{x}$ before multiplying? ::: Because the identity $\sqrt{a}\sqrt{b}=\sqrt{ab}$ is only guaranteed for non-negative reals; converting to $i$ first sidesteps the invalid rule and keeps the algebra honest. Why does multiplying by $i$ correspond to a $90^{\circ }$ rotation and not, say, $45^{\circ }$? ::: Because doing it twice must land on $-1$ (since $i^2=-1$), and $-1$ sits $180^{\circ }$ from $1$; two equal turns making $180^{\circ }$ means each turn is $90^{\circ }$. Why did mathematicians "invent" $i$ instead of declaring $x^2=-1$ unsolvable? ::: The same pattern created negatives (to solve $x+1=0$) and fractions (to solve $2x=1$): inventing a new number to solve a previously-impossible equation extends mathematics consistently, and it turned out to be enormously useful. Why is $1/i$ equal to $-i$ rather than something with $i$ in the denominator? ::: Multiplying top and bottom by $i$ clears the imaginary denominator: $\frac{1}{i}\cdot \frac{i}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i$. This "rationalising" trick always turns a division by $i$ into a clean number.

What is $i^0$, and does the usual "anything to the 0 is 1" survive here? ::: $i^0=1$. Yes — the rule holds for every nonzero base, and $i\ne 0$, so imaginariness is irrelevant. What is $i$ raised to a negative exponent, e.g. $i^{-3}$? ::: The cycle runs backwards: $i^{-1}=-i,i^{-2}=-1,i^{-3}=i,i^{-4}=1$. Formally $i^{-3}=1/i^3=1/(-i)=i$. Is $0$ a complex number? What are its real and imaginary parts? ::: Yes: $0=0+0i$. Both parts are $0$. Zero is simultaneously real, imaginary (trivially), and complex. Is the number $5i$ (pure imaginary) also a "complex number"? ::: Yes: $5i=0+5i$, a complex number with real part $0$. "Pure imaginary" just means the real part is zero. What happens to the sum of powers if you take five consecutive powers instead of four? ::: The four-term block sums to $0$, so five consecutive powers sum to just the fifth value. E.g. $i^1+\cdots +i^5=(i-1-i+1)+i=i$. Does $\sqrt{-a}\cdot \sqrt{-a}$ give $-a$ or $+a$ for positive $a$? ::: It gives $-a$: $\sqrt{-a}\cdot \sqrt{-a}=(i\sqrt{a})(i\sqrt{a})=i^{2}a=-a$. The naive $\sqrt{(-a)(-a)}=\sqrt{a^2}=a$ is the trap. If some real number $x$ satisfied $x^2=-1$, what contradiction follows? ::: Squaring any real gives $x^2\ge 0$, but $-1<0$. So $x^2\ge 0$ and $x^2=-1$ cannot both hold — no real $x$ exists, which is exactly why $i$ had to be invented.