Question

2.4 to 2.7 quiz
be sure to show work and circle/box your final answer.
for #s 1 - 6, simplify each:

1. $i^{33}$
2. $i^{400}$
3. $6i^{23}$
4. $sqrt{-81}$
5. $sqrt{-33}$
6. $sqrt{-45}$

for #s 7 - 16, write each expression as a complex number in standard form (a + bi).

1. $(2 - i)+(3 - 8i)$
2. $(9 + 4i)-(7 - 13i)$
3. $4i+(-8 - 2i)$
4. $13-(5 + 6i)$

Explanation:

Step1: Recall the powers of $i$

The powers of the imaginary unit $i$ have a cycle: $i^1 = i$, $i^2=- 1$, $i^3 = i^2\cdot i=-i$, $i^4=(i^2)^2 = 1$. For $i^{33}$, divide the exponent by 4. $33\div4 = 8$ remainder 1. So $i^{33}=(i^4)^8\cdot i^1$. Since $i^4 = 1$, then $i^{33}=i$.

Step2: For $i^{400}$

Divide 400 by 4. $400\div4=100$ with no remainder. So $i^{400}=(i^4)^{100}$. Since $i^4 = 1$, then $i^{400}=1$.

Step3: For $6i^{23}$

Divide 23 by 4. $23\div4 = 5$ remainder 3. So $i^{23}=(i^4)^5\cdot i^3$. Since $i^4 = 1$ and $i^3=-i$, then $6i^{23}=6\times(-i)=-6i$.

Step4: For $\sqrt{-81}$

We know that $\sqrt{-81}=\sqrt{81}\times\sqrt{-1}$. Since $\sqrt{81} = 9$ and $\sqrt{-1}=i$, then $\sqrt{-81}=9i$.

Step5: For $\sqrt{-33}$

We can write $\sqrt{-33}=\sqrt{33}\times\sqrt{-1}=\sqrt{33}i$.

Step6: For $\sqrt{-45}$

First, factor 45: $45 = 9\times5$. Then $\sqrt{-45}=\sqrt{45}\times\sqrt{-1}=\sqrt{9\times5}\times i = 3\sqrt{5}i$.

Step7: For $(2 - i)+(3 - 8i)$

Combine the real - parts and the imaginary - parts separately. $(2 + 3)+(-1-8)i=5 - 9i$.

Step8: For $(9 + 4i)-(7 - 13i)$

Distribute the negative sign: $9 + 4i-7 + 13i=(9 - 7)+(4 + 13)i=2+17i$.

Step9: For $4i+(-8 - 2i)$

Combine the real - parts and the imaginary - parts: $-8+(4 - 2)i=-8 + 2i$.

Step10: For $13-(5 + 6i)$

Distribute the negative sign: $13-5-6i = 8-6i$.

Answer:

1. $i$
2. $1$
3. $-6i$
4. $9i$
5. $\sqrt{33}i$
6. $3\sqrt{5}i$
7. $5 - 9i$
8. $2 + 17i$
9. $-8 + 2i$
10. $8-6i$