Use Euler's Formula to write 2sq.rt.2 -2isq.rt.2 in exponential form.
My answer: 4e^i(7pi/4)
I just need confirmation that my answer is correct. Thanks.
No, this is wrong, too.
It is
$\displaystyle a+ib = r*e^{i*\phi},$
where $\displaystyle r^2 = a^2+b^2 $
in this case: $\displaystyle r^2 = (2\sqrt{2})^2+ (2\sqrt{2})^2= 4*2+4*2 = 16$
=> $\displaystyle r = 4$
and $\displaystyle \phi = -arccos(a/r)$ (because a > 0, b <0)
in this case $\displaystyle \phi = -arccos(1) = -pi/4$
=> $\displaystyle 2\sqrt{2}-2i \sqrt{2} = 4e^{i*(-pi/4)} = 4cos(- pi/4) + 4sin(-pi/4)i$
Yours,
Rapha