# Check the correct sign $\pm$ in $( Ve^{-j(wt+A)})/(Ie^{-j(wt+B)})=(V/I)\cdot[\cos(A-B) \pm j\sin(A-B)]$

If $$v = Ve^{-j(wt+A)}, i= Ie^{-j(wt+B)}$$ then show that the impedance $$z = v/i$$ is given by $$Z = (V/I)\cdot[\cos(A-B) + j\sin(A-B)]$$

I get confused because when I used the correct rules, it always comes to $$(V/I)\cdot[\cos(A-B) - j\sin(A-B)]$$ instead because of the rule that $$e^{-j(\theta)}=\cos(\theta) - \sin(\theta)$$

• If $j=\sqrt{-1}$, your answer is correct (although you might want to check the rule $e^{-j(\theta)}=\cos(\theta) - \sin(\theta)$ again) – Shubham Johri Dec 16 '18 at 10:30

$$\frac{v}{i} = \frac{V}{I}e^{-j(A-B)} = \frac{V}{I} [\cos(-(A-B))+j\sin(-(A-B))]$$ by Euler's formula, then because cosine is even and sine is odd, $$\frac{v}{i} = \frac{V}{I} [\cos(A-B) -j \sin(A-B)].$$