True/False: If $(a+bi)^3 = 8$, then $a^2+b^2=4$ (yes it's missing the i in b^2)

I believe this is true because if take the cube root of both sides you get $a+bi=2$, squaring both sides will give you $a^2+b^2=2^2=4$

We also know that $a^2+b^2=z*\bar{z}=(a+bi)(a-bi)=a^2-(b^2*i^2) = a^2 - (b^2*(-1))= a^2+b^2$

True/False: If $\operatorname{Arg}(z)=\frac{3\pi}{4}$ and $\operatorname{Arg}(w)=\frac{-\pi}{2}$ then $\operatorname{Arg}(\frac{z}{w}) = \frac{5\pi}{4}$

I believe this is true because when dividing two complex numbers, you divide two complex numbers together you divide their magnitudes and subtract their angles. $\frac{3\pi}{4}-(-\frac{\pi}{2})=\frac{5\pi}{4}$

The reason why I'm asking is because my solution sheet is telling me these are both false but not why, and I think it's wrong.

  • $\begingroup$ $(a+bi)^2 = a^2+2abi-b^2 \ne a^2+b^2$? $\endgroup$ – ArsenBerk Aug 18 at 23:28
  • $\begingroup$ You certainly cannot say that $a+bi=2$, because there are two other cube roots of $8$. But take magnitudes first, and then you get $|a+bi|^3 = 8$, and so $|a+bi| = 2$ (why?). For the second one, how is $\text{Arg}$ defined? Maybe a picture might help. $\endgroup$ – Ted Shifrin Aug 18 at 23:29
  • $\begingroup$ I was thinking to multiply two complex numbers together as turning that same vector three times by the original arg(z), but even before that the big red flag i guess that should be going off is that 5$\pi$/4 isn't in between (-$\pi,\pi$]. $\endgroup$ – Krio Aug 18 at 23:55
  • $\begingroup$ @TedShifrin Correct my thinking, so if we're taking a cube of a complex number, for example here for question 1 and end up at a real number there are possibilities where (a+bi) is either a real or complex number, if it's a complex it must have 3 roots, each of those 3 roots have different arguments associated with them. BUT their magnitudes will all be the same which will be the cube root(magnitude of the cube)? $\endgroup$ – Krio Aug 19 at 0:07
  • $\begingroup$ Yes, they all have the same magnitude, because $|zw| = |z||w|$. So $|z^2| = |z|^2$, and $|z^3| = |z|^3$. $\endgroup$ – Ted Shifrin Aug 19 at 0:16

If you are told that $a$ and $b$ are real numbers then it is true that $a^{2}+b^{2}=4$. Perhaps you are not supposed to assume that these are real.

For the second question the answer depends on how arguments are defined. If you define argument as a number in $[-\pi, \pi)$ then we cannot have a complex number with argument $\frac {5\pi} 4$.

  • $\begingroup$ Ahh I missed the part that there was a restriction placed for arg(z) for us it's defined as (-pi,pi] $\endgroup$ – Krio Aug 18 at 23:47

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