What is (4-t+it-2i)^2(-1+i) ?

The solution to (4-t+it-2i)^2(-1+i) is (2t^2-8t+4)+(28+2t^2-16t)i

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simplify (4-t+it-2i)^2(-1+i)

simplify

(4 - t + i t - 2 i)^{2} (- 1 + i)

(2 t^{2} - 8 t + 4) + (28 + 2 t^{2} - 16 t) i

Solution steps

(4 - t + i t - 2 i)^{2} (- 1 + i)

Apply the distributive law:

a (b + c) = ab + a c

(4 - t + i t - 2 i)^{2} (- 1 + i) = (4 - t + i t - 2 i)^{2} (- 1) + (4 - t + i t - 2 i)^{2} i

= (4 - t + i t - 2 i)^{2} (- 1) + (4 - t + i t - 2 i)^{2} i

Simplify

(4 - t + i t - 2 i)^{2} (- 1) + (4 - t + i t - 2 i)^{2} i : 28 i + 2 t^{2} - 8 t + 2 t^{2} i - 16 t i + 4

= 28 i + 2 t^{2} - 8 t + 2 t^{2} i - 16 t i + 4

Rewrite in standard complex form:

(2 t^{2} - 8 t + 4) + (28 + 2 t^{2} - 16 t) i

= (2 t^{2} - 8 t + 4) + (28 + 2 t^{2} - 16 t) i

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6 (\frac{3}{2} + 3 u)

What is (4-t+it-2i)^2(-1+i) ?
The solution to (4-t+it-2i)^2(-1+i) is (2t^2-8t+4)+(28+2t^2-16t)i