# how is write (1+i square root 3) ^ 5 in complex number?

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You need to write the expanded binomial as the standard form of complex number such that: `(1+i sqrt3)^5 = x + iy.`

``Raising the binomial to square yields:

`(1+i sqrt3)^2 = 1 + 2isqrt3 + i^2*3`

Using complex number theory yields: `i^2 = -1` .

`(1+i sqrt3)^2 = 1 + 2isqrt3 - 3 =gt (1+i sqrt3)^2 =-2 + 2isqrt3 `

You need to multiply `(1 + isqrt3)^2` by `(1+i sqrt3)^2` such that:

`(1+i sqrt3)^2*(1+i sqrt3)^2 = (1+i sqrt3)^4`

`(-2 + 2isqrt3)*(-2 + 2isqrt3) = (-2 + 2isqrt3)^2`

`(-2 + 2isqrt3)*(-2 + 2isqrt3) = 4 - 8isqrt3 + 12i^2`

`(-2 + 2isqrt3)*(-2 + 2isqrt3) = 4 - 8isqrt3 - 12=gt(-2 + 2isqrt3)^2 =-8 - 8isqrt3`

Hence `(1+i sqrt3)^4 =-8 - 8isqrt3` .

You need to multiply `(1+i sqrt3)^4` by `(1+i sqrt3)` such that:

`(1+i sqrt3)^4*(1+i sqrt3) = (1+i sqrt3)^5`

Plugging `-8 - 8isqrt3` instead of `(1+i sqrt3)^4` in multiplication yields:

`(-8 - 8isqrt3)*(1+i sqrt3) = -8-8isqrt3 - 8isqrt3 - 24i^2`

`(-8 - 8isqrt3)*(1+i sqrt3) = 16 - 16i*sqrt3`

**Hence, writing the given binomial `(1+i sqrt3)^5` as the standard form of complex number yields `(1+i sqrt3)^5` = `16 - 16i*sqrt3` .**