Answer:
 d) √2∠(7π/4)
Step-by-step explanation:
Sometimes the expression A(cos(α)+i·sin(α)) is written as A·cis(α). I prefer the notation A∠α, because it is more compact.
You are given ...
 w = √2∠(π/4)
 z = 2∠(π/2)
And you are asked to find the difference w-z in polar form.
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In rectangular form, ...
 w = √2∠(π/4) = √2(1/√2 +i/√2) = 1 +i
 z = 2∠(π/2) = 2(0 +i·1) = 2i
Then the difference is ...
 w -z = (1 +i) -(2i) = 1 -i
You may notice this is the conjugate of w, so will have the opposite angle:
 w - z = w* = √2∠(-π/4) = √2∠(7π/4)
In the terms used by the problem statement, ...
 w -z = √2(cos(7π/4) +i·sin(7π/4)) . . . . matches choice D
