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I am having some trouble with a mathematical induction proof.

The question is the following:

Prove by mathematical induction that $\vert z_1 \cdot z_2 \cdot z_3 \cdots z_n \vert = \vert z_1 \vert \vert z_2 \vert \vert z_3 \vert \cdots \vert z_n \vert$

I've read through some mathematical induction material and, as far as I could grasp it, I needed to do the following:

1 - Check if the statement is true for $n = 1$

For $n=1$, $\vert z_1 \vert = \vert z_1 \vert$ (which is true)

2 - Build the case $n = k$ and assume that it is true

$\vert z_1 \cdot z_2 \cdot z_3 \cdots z_k \vert = \vert z_1 \vert \vert z_2 \vert \vert z_3 \vert \cdots \vert z_k \vert$

3 - Consider $n = k+1$ and check its validity (this is where things get unclear)

$\vert z_1 \cdot z_2 \cdot z_3 \cdots z_k \cdot z_{k+1} \vert = \vert z_1 \vert \vert z_2 \vert \vert z_3 \vert \cdots \vert z_k \vert \cdot \vert z_{k+1} \vert$

I could now substitute the previous expression and get

$\vert z_1 \cdot z_2 \cdot z_3 \cdots z_k \cdot z_{k+1} \vert = \vert z_1 \cdot z_2 \cdot z_3 \cdots z_k \vert \cdot \vert z_{k+1} \vert$

But I'm not sure this is the way to go. Any hint/help is highly appreciated.

bru1987
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3 Answers3

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In the case $n=k+1$, we have: \begin{align} |z_1 \cdots z_{k+1}| &= |(z_1 \cdots z_k) \cdot z_{k+1}| \\ &= |z_1 \cdots z_k| \cdot |z_{k+1}| \tag{$*$}\\ &= (|z_1| \cdots |z_k|) \cdot |z_{k+1}| \tag{induction hypothesis}\\ &= |z_1| \cdots |z_{k+1}| \end{align} Where throughout, I've freely used basic things like associativity of multiplication. Now, one thing which you have to prove is that $|a \cdot b| = |a| \cdot |b|$; this is what I made use of in $(*)$, with $a= z_1 \cdots z_k$ and $b= z_{k+1}$. This is a standard proof using the definition of absolute value of a complex number, so I'll omit it.

peek-a-boo
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You are missing a piece of information to prove this. Let $$z = x+iy$$ $$|z| = \sqrt{x^2+y^2}$$ so $$|z_1 * z_2| = |x_1 x_2 - y_1 y_2 + i(x_1y_2 + x_2 y_1)|$$ $$ = \sqrt{(x_1 x_2 - y_1 y_2)^2 + (x_1y_2 + x_2 y_1)^2}$$ $$ = \sqrt{(x_1 x_2)^2 - 2x_1 x_2 y_1 y_2 + (y_1 y_2)^2 + (x_1 y_2)^2 + 2x_1x_2y_1y_2 + (x_2 y_1)^2}$$ $$ = \sqrt{(x_1 x_2)^2 + (x_1 y_2)^2 + (x_2 y_1)^2 + (y_1 y_2)^2}$$ $$ = \sqrt{(x_1^2 + y_1^2)(x_2^2 + y_2^2)} = \sqrt{(x_1^2 + y_1^2)}\sqrt{(x_2^2 + y_2^2)}$$ $$ = |z_1||z_2|$$ and your result follows from the fact that the complex numbers are closed under multiplication, i.e. $$|z_1 * z_2 * \ldots * z_k * z_{k+1}| = |(z_1 * z_2 * \ldots * z_k) * z_{k+1}| = |z_1 * z_2 * \ldots * z_k| * |z_{k+1}|$$ $$= |z_1| * |z_2| * \ldots * |z_{k+1}|$$ by the inductive hypothesis. Disregard the statement I made about induction with complex numbers before.

Ryan Shesler
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Comments on Induction: It's worthwhile noting that when you write $|z_{1}\cdot z_{2}\cdots z_{k}\cdot z_{k+1}|=|z_{1}\cdot z_{2}\cdots z_{k}\cdot||z_{k+1}|,$ you're actually using the case $|xy|=|x||y|,$ where $x=z_{1}\cdot z_{2}\cdots z_{k}$ and $ y = z_{k+1}$. Can you see where you have to use the assumption now?

I think that the hardest part to grasp from induction is precisely when to use the assumption (called the hypothesis of induction). Try this one:

If a $1\times 1$ square contains infinite points, then for any natural number $n$, there is a $\displaystyle \frac{1}{2^{n}}\times \frac{1}{2^{n}}$ square inside it that also has infinite points.

Part of the solution:

Base case: cut the square into 4 quarters to create four $\displaystyle \frac{1}{2}\times \frac{1}{2}$ squares. Then at least one of those must have infinite points because otherwise, all of them would have finitely many and that's the whole square.

Hypothesis of Induction: Suppose it's true for some $n$. Meaning, we have managed to find a $\displaystyle \frac{1}{2^{n}}\times \frac{1}{2^{n}}$ square with infinite points inside the original.

How would you get a $ \displaystyle \frac{1}{2^{n+1}}\times \frac{1}{2^{n+1}}$ square? How many of those squares will you get? Why would at least one of them have infinite points? What allows us to say that such infinite number of points exist? How would this help us prove what we wanted to prove?

Jay
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