Question

2.2 quiz - complex operations
simplify. write all answers in the form ( a + bi ) and put a box or circle around ( a ).
when multiplying and dividing you must show your work for full credit.

1. ( (-6 - 8i)+(8 - 2i) ) 2) ( (2 - i)+(4 + 3i) )

a) ( 6 - 4i ) b) ( -2 + 2i )
c) ( 6 + 2i ) d) ( 8 - i )

1. ( (-7 + 8i)+(3 - 3i) ) 4) ( (-8 + 3i)-(-5 - 6i) )

a) ( -3 - 3i ) b) ( 15i )
c) ( -3 + 9i ) d) ( -3 + 3i )

1. ( (-5 + 5i)-(2 + 5i) ) 6) ( (-5 - 5i)-(7 + 2i) )

a) ( 2 + 3i ) b) ( -2 + 3i )
c) ( -12 - 7i ) d) ( -2 - 7i )

1. ( (2 - 3i)(-2 + 5i) ) 8) ( (-5 + 2i)(-1 - 2i) )

a) ( -1 - 12i ) b) ( -9 - 8i )
c) ( 9 + 8i ) d) ( 1 - 12i )

1. ( \frac{5i}{4 + i} ) 10) ( \frac{6 + 4i}{4 + 6i} )

a) ( \frac{7}{10}-\frac{5i}{21} ) b) ( \frac{30}{37}-\frac{32i}{37} )
c) ( \frac{12}{13}-\frac{5i}{13} ) d) ( 1-\frac{i}{2} )

Explanation:

Step1: Combine real and imaginary parts

For \((-6 - 8i)+(8 - 2i)\), we have \((-6 + 8)+(-8i-2i)\).

Step2: Calculate

\(-6 + 8 = 2\) and \(-8i-2i=-10i\). So the result is \(2-10i\).

Step3: For \((12 - i)+(4 + 3i)\)

Combine real and imaginary parts: \((12 + 4)+(-i + 3i)\).

Step4: Calculate

\(12+4 = 16\) and \(-i+3i = 2i\). But wait, let's check the options. Wait, no, for \((12 - i)+(4 + 3i)\) it's \(16 + 2i\) which is not in the options. Wait, maybe a typo. Wait, no, for \((-7 + 8i)+(3 - 3i)\)

Step5: Combine real and imaginary parts

\((-7+3)+(8i-3i)\)

Step6: Calculate

\(-7 + 3=-4\) and \(8i-3i = 5i\).

Step7: For \((-8 + 3i)-(-5 - 6i)\)

Distribute the negative sign: \(-8 + 3i + 5+6i\).

Step8: Combine real and imaginary parts

\((-8 + 5)+(3i+6i)\)

Step9: Calculate

\(-8 + 5=-3\) and \(3i+6i = 9i\). So it's \(-3 + 9i\) (Option C).

Step10: For \((-5 + 5i)-(2 + 5i)\)

Distribute the negative sign: \(-5 + 5i-2-5i\).

Step11: Combine real and imaginary parts

\((-5-2)+(5i-5i)\)

Step12: Calculate

\(-5-2=-7\) and \(5i - 5i=0\). So it's \(-7\).

Step13: For \((-5 - 5i)-(7 + 2i)\)

Distribute the negative sign: \(-5-5i-7-2i\).

Step14: Combine real and imaginary parts

\((-5-7)+(-5i-2i)\)

Step15: Calculate

\(-5-7=-12\) and \(-5i-2i=-7i\). So it's \(-12-7i\) (Option C).

Step16: For \((2 - 3i)(-2 + 5i)\)

Use FOIL: \(2\times(-2)+2\times5i-3i\times(-2)-3i\times5i\).
\(=-4 + 10i + 6i-15i^{2}\). Since \(i^{2}=-1\), it becomes \(-4+16i + 15=11 + 16i\).

Step17: For \((-5 + 2i)(-1 - 2i)\)

Use FOIL: \((-5)\times(-1)+(-5)\times(-2i)+2i\times(-1)+2i\times(-2i)\).
\(=5 + 10i-2i-4i^{2}\). Since \(i^{2}=-1\), it becomes \(5 + 8i+4=9 + 8i\) (Option C).

Step18: For \(\frac{5i}{4 + i}\)

Multiply numerator and denominator by the conjugate \(4 - i\): \(\frac{5i(4 - i)}{(4 + i)(4 - i)}\).
\(=\frac{20i-5i^{2}}{16 - i^{2}}\). Since \(i^{2}=-1\), it's \(\frac{20i + 5}{17}=\frac{5}{17}+\frac{20}{17}i\).

Step19: For \(\frac{6 + 4i}{4 + 6i}\)

Multiply numerator and denominator by the conjugate \(4 - 6i\): \(\frac{(6 + 4i)(4 - 6i)}{(4 + 6i)(4 - 6i)}\).
\(=\frac{24-36i+16i-24i^{2}}{16 - 36i^{2}}\). Since \(i^{2}=-1\), \(\frac{24-20i + 24}{16 + 36}=\frac{48-20i}{52}=\frac{12}{13}-\frac{5i}{13}\) (Option C).

Answer:

1. \(2-10i\)
2. (There might be a typo in the problem statement as the correct result of \((12 - i)+(4 + 3i)=16 + 2i\) is not in the options. Assuming it was \((2 - i)+(4 + 3i)\) then \(6 + 2i\) (Option C))
3. \(-4 + 5i\)
4. C. \(-3 + 9i\)
5. \(-7\)
6. C. \(-12-7i\)
7. \(11 + 16i\)
8. C. \(9 + 8i\)
9. \(\frac{5}{17}+\frac{20}{17}i\)
10. C. \(\frac{12}{13}-\frac{5i}{13}\)