Before we embark on further explorations with our complex number $i$i, let's have a look at what happens when we take successive powers of $i$i.
Try this activity yourself first, before checking out my solution.
Make a list of powers of $i$i, up to $i^{15}$i15.
Simplify the results.
Then generalise the pattern.
(see here for the solution)
Now that you have explored how to simplify powers of $i$i, the only other thing to do is combine this with other algebraic simplifications.
Simplify $(2i)^4$(2i)4
We need to remember our index laws here, so $\left(ab\right)^n=a^n\times b^n$(ab)n=an×bn
Thus,
$\left(2\times1\right)^4$(2×1)4 | $=$= | $2^4\times i^4$24×i4 |
$=$= | $16\times i^2\times i^2$16×i2×i2 | |
$=$= | $16\times\left(-1\right)\times\left(-1\right)$16×(−1)×(−1) | |
$=$= | $16$16 |
Simplify $4i^2-3i^3-7i^4$4i2−3i3−7i4
$4i^2-3i^3-7i^4$4i2−3i3−7i4 | $=$= | $4\times-1-3\times-i-7\times1$4×−1−3×−i−7×1 |
$=$= | $-4--3i-7$−4−−3i−7 | |
$=$= | $3i-11$3i−11 |
Simplify $2i^7$2i7.
Simplify $\left(2i\right)^9$(2i)9.
Simplify $\left(\sqrt{5}i\right)^6$(√5i)6.