Show by example that it is not always true that arg(ab)=arg…

Show by example that it is not always true that $\arg(ab)=\arg a+\arg b$ when using the principal value determination $-\pi<\arg z\leq\pi$ $(z\in\mathbb{C}^*)$.

Step 1 · Choose the example

Let $a=b=-1+i$.

Step 2 · Compute $\arg a$ and $\arg b$

$-1+i$ lies in the second quadrant with $|a|=\sqrt2$; its reference angle is $\frac\pi4$, so:

$$ \arg a=\arg b=\pi-\frac\pi4=\frac{3\pi}4 $$

Step 3 · Sum of the principal arguments

$$ \arg a+\arg b=\frac{3\pi}4+\frac{3\pi}4=\frac{3\pi}2 $$

Since $\frac{3\pi}2>\pi$, this sum already lies outside the principal range $(-\pi,\pi]$.

Step 4 · Compute $ab$ and its principal argument

$$ ab=(-1+i)^2=1-2i+i^2=-2i $$

$-2i$ lies on the negative imaginary axis, so $\arg(ab)=-\dfrac\pi2$.

Step 5 · Compare

$$ \arg(ab)=-\frac\pi2 \qquad\neq\qquad \arg a+\arg b=\frac{3\pi}2 $$

but the two values differ by exactly $2\pi$:

$$ -\frac\pi2 = \frac{3\pi}2-2\pi $$

confirming $\boxed{\arg(ab)\equiv\arg a+\arg b\pmod{2\pi}}$, even though the principal values themselves do not add.

Step 6 · A second example

Let $a=-1$, $b=i$: $\arg a=\pi$, $\arg b=\frac\pi2$, so $\arg a+\arg b=\frac{3\pi}2$. But $ab=-i$, so $\arg(ab)=-\frac\pi2$. Again $-\frac\pi2=\frac{3\pi}2-2\pi$.

Step 7 · The general rule

For principal arguments, the correct statement is:

$$ \arg(ab)=\arg a+\arg b+2\pi k, \qquad k\in\{-1,0,1\} \text{ chosen so that the result lies in } (-\pi,\pi] $$