Question

name: date: period: sc directions: read each statement carefully. choose and circle the letter of the correct 1.) simplify \\( \sqrt { - 1 2 1 } \\) 2.) simplify \\( i ^ { 1 2 5 } \\) a.) - 1 1 i a.) - i b.) 1 1 i b.) 1 c.) 1 1 c.) - 1 d.) - 1 1 d.) i 3.) simplify the expression: 4.) for what values of a and b will \\( ( 3 - 2 i ) ( 4 + 5 i ) \\) the expression \\( ( 7 - 2 i ) ( a + b i ) \\) a.) 2 result in a real number? b.) 2 2 \\( a = \\) c.) \\( 2 + 7 i \\) \\( b = \\) d.) \\( 2 2 + 7 i \\) 5.) add \\( ( - 3 + 2 i ) + ( 5 - 6 i ) \\) 6.) subtract \\( ( 1 1 - 7 i ) - ( - 3 + 1 2 i ) \\) a.) - 2 + 4 i a.) 1 4 - 1 9 i b.) 2 - 3 i b.) 1 3 - 1 9 i c.) - 4 + 2 i c.) 8 + 5 i d.) 2 - 4 i d.) 8 - 1 9 i 7.) multiply \\( ( - 5 i ) ( i ) \\) 8.) simplify \\( 2 i ( 6 - i ) \\) a.) - 5 i a.) 1 1 i b.) - 5 b.) 1 2 - 2 i c.) 5 c.) 2 + 1 2 i d.) \\( 5 i ^ { 2 } \\) d.) - 2 + 1 2 i 9.) multiply \\( ( - 3 + 2 i ) ( 1 - 4 i ) \\) 10.) simplify \\( ( 5 - 4 i ) ^ { 2 } \\) a.) 5 + 1 4 i a.) 2 5 + 4 0 i b.) - 3 - 8 i b.) 4 1 c.) - 2 - 2 i c.) 9 - 4 0 i d.) - 1 1 - 1 0 i d.) 4 1 - 4 0 i

Explanation:

Step1: Simplify $\sqrt{-121}$

Use the property $\sqrt{-a}=\sqrt{a}i$ ($a>0$). Here $a = 121$, and $\sqrt{121}=11$. So $\sqrt{-121}=\sqrt{121}\times\sqrt{-1}=11i$.

Step2: Simplify $i^{125}$

Recall the pattern of powers of $i$: $i^1 = i$, $i^2=-1$, $i^3=-i$, $i^4 = 1$, and the cycle repeats every 4. Divide 125 by 4: $125\div4 = 31\cdots\cdots1$. So $i^{125}=(i^4)^{31}\times i^1$. Since $i^4 = 1$, then $(i^4)^{31}=1^{31}=1$, and $i^{125}=i$.

Step3: Simplify $(3 - 2i)(4 + 5i)$

Use the FOIL method: $(3-2i)(4 + 5i)=3\times4+3\times5i-2i\times4-2i\times5i=12 + 15i-8i-10i^2$. Since $i^2=-1$, we have $12 + 7i+10=22 + 7i$.

Step4: Simplify $(-3 + 2i)+(5 - 6i)$

Combine the real parts and the imaginary parts separately: $(-3 + 5)+(2i-6i)=2-4i$.

Step5: Simplify $(11 - 7i)-(-3 + 12i)$

Distribute the negative sign: $11-7i + 3-12i=(11 + 3)+(-7i-12i)=14-19i$.

Step6: Simplify $(-5i)(i)$

Multiply: $(-5i)(i)=-5i^2$. Since $i^2=-1$, then $-5i^2 = 5$.

Step7: Simplify $2i(6 - i)$

Use the distributive property: $2i\times6-2i\times i=12i-2i^2$. Since $i^2=-1$, we get $2 + 12i$.

Step8: Simplify $(-3 + 2i)(1 - 4i)$

Use the FOIL method: $(-3)\times1+(-3)\times(-4i)+2i\times1+2i\times(-4i)=-3 + 12i+2i-8i^2$. Since $i^2=-1$, we have $-3+14i + 8=5 + 14i$.

Step9: Simplify $(5 - 4i)^2$

Use the formula $(a - b)^2=a^2-2ab + b^2$. Here $a = 5$, $b = 4i$. So $(5 - 4i)^2=25-40i+16i^2$. Since $i^2=-1$, we get $25-40i-16=9-40i$.

Step10: Find $a$ and $b$ for $(7 - 2i)(a + bi)$ to be real

Expand $(7 - 2i)(a + bi)=7a+7bi-2ai-2bi^2=(7a + 2b)+(7b-2a)i$. For it to be real, the imaginary part must be 0, so $7b-2a = 0$, or $a=\frac{7}{2}b$. Let $b = 2$, then $a = 7$.

Answer:

1. b. \(11i\)
2. d. \(i\)
3. d. \(22 + 7i\)
4. \(a = 7\), \(b = 2\)
5. d. \(2-4i\)
6. a. \(14 - 19i\)
7. c. \(5\)
8. c. \(2 + 12i\)
9. a. \(5 + 14i\)
10. c. \(9 - 40i\)