# Complex Number True/False

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.

• $(a+bi)^2 = a^2+2abi-b^2 \ne a^2+b^2$? – ArsenBerk Aug 18 at 23:28
• 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. – Ted Shifrin Aug 18 at 23:29
• 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 (-$\pi,\pi$]. – Krio Aug 18 at 23:55
• @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)? – Krio Aug 19 at 0:07
• Yes, they all have the same magnitude, because $|zw| = |z||w|$. So $|z^2| = |z|^2$, and $|z^3| = |z|^3$. – 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 $$\frac {5\pi} 4$$.