Understanding the Phazor Equation: I_s(t)=sin(t)

[electron2]

for this signal
$$I_s(t)=sin(t)$$
i have this equation
[itex] I_L(1-CL+jCR+jcz)=I_s<br /> [/tex]<br /> then the next line is<br /> $$I_L(1-CL+jCR+jcz)=j$$<br /> <br /> why they substitute j with I_s[/itex]

 

[electron2]

$$sin(t)=cos(\frac{\pi}{2}-t)=cos(t-\frac{\pi}{2})$$
the formula of the signal is
$$Is=Acos(\omega t+\phi)$$
then we transform it to the phasor representation formula
$$Is=Ae^{j\phi}$$
so we get
$$Is=1e^{-j\frac{\pi}{2}}$$
and when we look at this expression as oilers formula we get
the Is=-j

so why its written Is=j
?

 

[CEL]

$$sin(t)=cos(\frac{\pi}{2}-t)=cos(t-\frac{\pi}{2})$$
the formula of the signal is
$$Is=Acos(\omega t+\phi)$$
then we transform it to the phasor representation formula
$$Is=Ae^{j\phi}$$
so we get
$$Is=1e^{-j\frac{\pi}{2}}$$
and when we look at this expression as oilers formula we get
the Is=-j

so why its written Is=j
?

$$e^{j\theta} = cos(\theta) + j sin(\theta)$$
for $$\theta = \frac{\pi}{2}$$
$$e^{j\frac{\pi}{2}} = cos(\frac{\pi}{2}) + j sin(\frac{\pi}{2}) = 0 + j.1 = j$$