$(1-i)^{40}=2^a$

$(1-i)^{40} = ((1-i)^2)^{20}$

$= (1-2i - 1)^{20}$

$= (-2i)^{20}$

$-2i$ to the power of $20$ is equal to $(2i)^{20}$

So,

$(2i)^{20} = ((2i)^2)^{10}$

$=(4\cdot(-1))^{10}$

$=(-4)^{10}$

Same as before, $-4$ to the tenth power is same as $4$ to the tenth power

We can write $4^{10}$ to be $2^{20}$

Therefore, $a = \boxed{20}$

(1-i)^40=-2i[(1-i)^38]=(-2i)^20=[(-1)^20][2^20][i^20]=2^20. Therefore, a=20.

Using polar form, $(1 - i)^{40}$ has argument $\sqrt{1^2+1^2} = \sqrt2$ , and modulus $-\frac{\pi}{4}$ . Therefore, this equals $\left(\sqrt{2} e^{i -\pi/4}\right)^{40} = \sqrt{2}^{40} e^{-10 i \pi} = 2^{20}$ as $e^{2 i \pi} = 1$ .