Question

Tính z=(〖(1+i)〗^10-1)/(〖(1+i)〗^10+1)

Answer 1

It looks like the given complex number is

z = ((1 + i )¹⁰ - 1) / ((1 + i )¹⁰ + 1)

Let w = 1 + i, so we rewrite

z = (w ¹⁰ - 1) / (w ¹⁰ + 1)

Since w * = 1 - i, we get w w * = (1 + i ) (1 - i ) = 1 - (-1) = 2. Multiply z by ((w *)¹⁰ + 1) / ((w *)¹⁰ + 1). This gives

z = (2¹⁰ - 1 + w ¹⁰ - (w *)¹⁰) / (2¹⁰ + 1 + w ¹⁰ + (w *)¹⁰)

Now, writing w in polar form yields

w = √2 exp(i π/4)

==>   w ¹⁰ = (√2)¹⁰ exp(i 10π/4) = 2⁵ exp(i 5π/2) = 2⁵ i

Similarly,

w * = √2 exp(-i π/4)

==>   (w *)¹⁰ = -2⁵ i

So we have

w ¹⁰ + (w *)¹⁰ = 0

w ¹⁰ - (w *)¹⁰ = 2 × 2⁵ i = 2⁶ i

Then we end up with

z = (2¹⁰ - 1 + 2⁶ i ) / (2¹⁰ + 1) = (1023 + 64i ) / 1025