1. ## the right question

write in rectangular form

(square root 2-i) 4th power

2. $\left ( \sqrt{2-i} \right )^4$

= $(2-i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i$

-Dan

3. Hello, Shorty!

Is this what you meant?

Write in rectangular form: $(\sqrt{2} - i)^4$

Just expand it . . . what's stopping you?

$(\sqrt{2} - i)^4 \;=\;(\sqrt{2})^4 - 4(\sqrt{2})^3(i) + 6(\sqrt{2})^2(i^2) - 4(\sqrt{2})(i^3) + (i^4)$

. . . . . . . $= \;16 - 4(2\sqrt{2})(i) + 6(2)(-1) - 4(\sqrt{2})(-i) + 1$

. . . . . . . $= \;16 - 8\sqrt{2}i - 12 + 4\sqrt{2}i + 1$

. . . . . . . $= \;5 - 4\sqrt{2}i$