If x = 1 + i, then what is the value of x 6+ x 4+ x 2+ 1?
-6i - 3
We are given a complex number \(x = 1 + i\) and asked to find the value of the expression \(x^6 + x^4 + x^2 + 1\). To solve this, we need to calculate the powers of \(x\) individually and then substitute them into the expression.
First, let's find the value of \(x^2\):
\(x^2 = (1 + i)^2\)
Using the formula \((a+b)^2 = a^2 + 2ab + b^2\), where \(a=1\) and \(b=i\):
\(x^2 = 1^2 + 2(1)(i) + i^2\)
Recall that \(i^2 = -1\). Substituting this value:
\(x^2 = 1 + 2i - 1\)
\(x^2 = 2i\)
Next, let's find the value of \(x^4\). We can calculate this as \((x^2)^2\):
\(x^4 = (2i)^2\)
\(x^4 = 2^2 \cdot i^2\)
\(x^4 = 4 \cdot (-1)\)
\(x^4 = -4\)
Finally, let's find the value of \(x^6\). We can calculate this as \(x^2 \cdot x^4\):
\(x^6 = x^2 \cdot x^4\)
Substitute the values we found for \(x^2\) and \(x^4\):
\(x^6 = (2i) \cdot (-4)\)
\(x^6 = -8i\)
Now we substitute the calculated values of \(x^6\), \(x^4\), and \(x^2\) into the given expression:
\(x^6 + x^4 + x^2 + 1 = (-8i) + (-4) + (2i) + 1\)
Rearrange the terms to group the real parts and the imaginary parts:
\(x^6 + x^4 + x^2 + 1 = (-4 + 1) + (-8i + 2i)\)
Combine the real parts:
\(-4 + 1 = -3\)
Combine the imaginary parts:
\(-8i + 2i = -6i\)
Putting the real and imaginary parts together, we get:
\(x^6 + x^4 + x^2 + 1 = -3 - 6i\)
Thus, the value of the expression \(x^6 + x^4 + x^2 + 1\) when \(x = 1 + i\) is \(-3 - 6i\).
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