Step 1:
Let $1+i=r(\cos \theta+i\sin\theta)$
Equating the real imaginary parts separately
$r\cos\theta=1$
$r\sin\theta=1$
Squaring and adding we get
$r^2=2$
$\Rightarrow r=\sqrt 2$
Step 2:
Now $\alpha=\tan^{-1}1=\large\frac{\pi}{4}$
$\theta=\alpha=\large\frac{\pi}{4}$
Since $1+i$ corresponds to a point in quadrant 1.
Therefore $1+i=\sqrt 2(\cos\large\frac{\pi}{4}$$+i\sin\large\frac{\pi}{4})$
Step 3:
$(1+i)^{4n}=2^{\large\frac{4n}{2}}$$(\cos\large\frac{4n\pi}{4}$$+i\sin\large\frac{4n\pi}{4})$
$\qquad\;\;\quad=2^{2n}(\cos n\pi+i\sin n\pi)$
Now $\sin n\pi=0$ for $n\in N$
Therefore $(1+i)^{4n}$ is purely real.
Step 4:
$(1+i)^{4n+2}=2^{\large\frac{4n+2}{2}}$$(\cos\large\frac{(4n+2)\pi}{4}$$+i\sin\large\frac{(4n+2)\pi}{4})$
$\qquad\qquad=2^{2n+1}(\cos(n\pi+\large\frac{\pi}{2})$$+i\sin(n\pi+\large\frac{\pi}{2}))$
Now $\cos(n\pi+\large\frac{\pi}{2})$$=0$ for $n\in N$
Therefore $(1+i)^{4n+2}$ is purely imaginary.