I have another method :)
$$cos\theta+isin\theta = e^{i\theta}\\\\$$
where theta is in radians
$$\\148^0=\frac{148\pi}{180}\; radians\\\\
11^0=\frac{11\pi}{180}\; radians\\\\$$
Given Z = 2(cos 148º + isin 148º) and W = 5(cos 11º + isin 11º), find and simplify Z divided by W.
becomes
$$\\\dfrac{2e^{(148\pi/180)i}}{5e^{(11\pi/180)i}}\\\\
=0.4e^{[(148\pi/180)-(11\pi/180)]i}\\\\$$
$$\left({\frac{{\mathtt{148}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{180}}}}\right){\mathtt{\,-\,}}\left({\frac{{\mathtt{11}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{180}}}}\right) = {\mathtt{2.391\: \!101\: \!075\: \!232\: \!231\: \!5}}$$
=0.4cos (2.3911010752322315) +0.4* isin(2.3911010752322315)
remember this is in radians.
= -0.2925 + 0.2728i
= -0.29 + 0.27i
I think that method is correct.