QUESTION IMAGE
Question
name:
topic : addition and subtraction of complex numbers - worksheet 1
solve the problems. express the result in a + bi form
- (5 + 3i) - (2 - 2i)
- (8 + 3i) + (5 - 6i)
- (6 + 8i) - (3 + 5i)
- (4 - √(-16)) + (2 - √(-25))
- (6 - √(-64)) + (12 + √(-81))
- (4 - 8i) - (2 - 5i)
- find the sum 6 - 4i and 6 - 4i subtract 3 - 2i from it.
- subtract 8 - 6i from 9 + 3i
- add 12 + 5i, 8 + 6i and 8 - 6i
- subtract 15 - 6i from 13 - 4i
- (-2 + 3i) + (7 - 4i) - (4 + 5i)
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Let's solve these complex number problems one by one. We'll use the rule that for complex numbers \(a + bi\) and \(c + di\), addition is \((a + c)+(b + d)i\) and subtraction is \((a - c)+(b - d)i\), and \(\sqrt{-n}=i\sqrt{n}\) for positive real numbers \(n\).
Problem 1: \((5 + 3i)-(2 - 2i)\)
Step 1: Distribute the subtraction
We can rewrite the subtraction as addition of the negative: \((5 + 3i)+(-2 + 2i)\)
Step 2: Combine real and imaginary parts
For real parts: \(5-2 = 3\)
For imaginary parts: \(3i + 2i=5i\)
So the result is \(3 + 5i\)
Problem 2: \((8 + 3i)+(5 - 6i)\)
Step 1: Combine real parts
\(8 + 5=13\)
Step 2: Combine imaginary parts
\(3i-6i=-3i\)
So the result is \(13-3i\)
Problem 3: \((6 + 8i)-(3 + 5i)\)
Step 1: Distribute the subtraction
\((6 + 8i)+(-3 - 5i)\)
Step 2: Combine real and imaginary parts
Real parts: \(6-3 = 3\)
Imaginary parts: \(8i-5i = 3i\)
Result: \(3 + 3i\)
Problem 4: \((4-\sqrt{-16})+(2-\sqrt{-25})\)
Step 1: Simplify square roots of negative numbers
We know that \(\sqrt{-16}=i\sqrt{16} = 4i\) and \(\sqrt{-25}=i\sqrt{25}=5i\)
So the expression becomes \((4 - 4i)+(2 - 5i)\)
Step 2: Combine real and imaginary parts
Real parts: \(4 + 2=6\)
Imaginary parts: \(-4i-5i=-9i\)
Result: \(6-9i\)
Problem 5: \((6-\sqrt{-64})+(12+\sqrt{-81})\)
Step 1: Simplify square roots of negative numbers
\(\sqrt{-64}=i\sqrt{64} = 8i\) and \(\sqrt{-81}=i\sqrt{81}=9i\)
The expression becomes \((6 - 8i)+(12 + 9i)\)
Step 2: Combine real and imaginary parts
Real parts: \(6+12 = 18\)
Imaginary parts: \(-8i + 9i=i\)
Result: \(18 + i\)
Problem 6: \((4 - 8i)-(2 - 5i)\)
Step 1: Distribute the subtraction
\((4 - 8i)+(-2 + 5i)\)
Step 2: Combine real and imaginary parts
Real parts: \(4-2 = 2\)
Imaginary parts: \(-8i+5i=-3i\)
Result: \(2-3i\)
Problem 7: Find the sum of \(6 - 4i\) and \(6 - 4i\), then subtract \(3 - 2i\) from it.
Step 1: Sum of \(6 - 4i\) and \(6 - 4i\)
\((6 - 4i)+(6 - 4i)=12-8i\)
Step 2: Subtract \(3 - 2i\) from \(12-8i\)
\((12-8i)-(3 - 2i)=(12-3)+(-8i + 2i)=9-6i\)
Problem 8: Subtract \(8 - 6i\) from \(9 + 3i\)
Step 1: Rewrite the subtraction
\((9 + 3i)-(8 - 6i)=(9-8)+(3i + 6i)\)
Step 2: Combine real and imaginary parts
Real parts: \(9-8 = 1\)
Imaginary parts: \(3i+6i = 9i\)? Wait, no: \(3i-(-6i)=3i + 6i=9i\)? Wait, original subtraction: \((9 + 3i)+(-8 + 6i)\)
Real parts: \(9-8 = 1\)
Imaginary parts: \(3i+6i = 9i\)? Wait, no, the problem says "Subtract \(8 - 6i\) from \(9 + 3i\)", so it is \((9 + 3i)-(8 - 6i)=9 + 3i-8 + 6i=(9-8)+(3i + 6i)=1 + 9i\)? Wait, I made a mistake earlier. Let's correct:
\((9 + 3i)-(8 - 6i)=9-8+3i + 6i=1 + 9i\)
Problem 9: Add \(12 + 5i\), \(8 + 6i\) and \(8 - 6i\)
Step 1: Add the first two complex numbers
\((12 + 5i)+(8 + 6i)=20 + 11i\)
Step 2: Add the third complex number
\((20 + 11i)+(8 - 6i)=20 + 8+11i-6i=28 + 5i\)
Problem 10: Subtract \(15 - 6i\) from \(13 - 4i\)
Step 1: Rewrite the subtraction
\((13 - 4i)-(15 - 6i)=13 - 4i-15 + 6i\)
Step 2: Combine real and imaginary parts
Real parts: \(13-15=-2\)
Imaginary parts: \(-4i + 6i = 2i\)
Result: \(-2 + 2i\)
Problem 11: \((-2 + 3i)+(7 - 4i)-(4 + 5i)\)
Step 1: First, add the first two complex numbers
\((-2 + 3i)+(7 - 4i)=(-2 + 7)+(3i-4i)=5 - i\)
Step 2: Subtract the third complex number
\((5 - i)-(4 + 5i)=5 - i-4 - 5i=(5-4)+(-i-5i)=1-6i\)
Final Answers:
- \(\boldsymbol{3 + 5i}\)
- \(\boldsymbol{13-3i}\)
- \(\boldsymbol{3 + 3i}\)
- \(\boldsymbol{6-9i}\)
- \(\boldsymbol{18 + i}\)
- \(\boldsymbol{2-3i}\)
- \(\boldsymbol{9-6i}\) (co…
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Let's solve these complex number problems one by one. We'll use the rule that for complex numbers \(a + bi\) and \(c + di\), addition is \((a + c)+(b + d)i\) and subtraction is \((a - c)+(b - d)i\), and \(\sqrt{-n}=i\sqrt{n}\) for positive real numbers \(n\).
Problem 1: \((5 + 3i)-(2 - 2i)\)
Step 1: Distribute the subtraction
We can rewrite the subtraction as addition of the negative: \((5 + 3i)+(-2 + 2i)\)
Step 2: Combine real and imaginary parts
For real parts: \(5-2 = 3\)
For imaginary parts: \(3i + 2i=5i\)
So the result is \(3 + 5i\)
Problem 2: \((8 + 3i)+(5 - 6i)\)
Step 1: Combine real parts
\(8 + 5=13\)
Step 2: Combine imaginary parts
\(3i-6i=-3i\)
So the result is \(13-3i\)
Problem 3: \((6 + 8i)-(3 + 5i)\)
Step 1: Distribute the subtraction
\((6 + 8i)+(-3 - 5i)\)
Step 2: Combine real and imaginary parts
Real parts: \(6-3 = 3\)
Imaginary parts: \(8i-5i = 3i\)
Result: \(3 + 3i\)
Problem 4: \((4-\sqrt{-16})+(2-\sqrt{-25})\)
Step 1: Simplify square roots of negative numbers
We know that \(\sqrt{-16}=i\sqrt{16} = 4i\) and \(\sqrt{-25}=i\sqrt{25}=5i\)
So the expression becomes \((4 - 4i)+(2 - 5i)\)
Step 2: Combine real and imaginary parts
Real parts: \(4 + 2=6\)
Imaginary parts: \(-4i-5i=-9i\)
Result: \(6-9i\)
Problem 5: \((6-\sqrt{-64})+(12+\sqrt{-81})\)
Step 1: Simplify square roots of negative numbers
\(\sqrt{-64}=i\sqrt{64} = 8i\) and \(\sqrt{-81}=i\sqrt{81}=9i\)
The expression becomes \((6 - 8i)+(12 + 9i)\)
Step 2: Combine real and imaginary parts
Real parts: \(6+12 = 18\)
Imaginary parts: \(-8i + 9i=i\)
Result: \(18 + i\)
Problem 6: \((4 - 8i)-(2 - 5i)\)
Step 1: Distribute the subtraction
\((4 - 8i)+(-2 + 5i)\)
Step 2: Combine real and imaginary parts
Real parts: \(4-2 = 2\)
Imaginary parts: \(-8i+5i=-3i\)
Result: \(2-3i\)
Problem 7: Find the sum of \(6 - 4i\) and \(6 - 4i\), then subtract \(3 - 2i\) from it.
Step 1: Sum of \(6 - 4i\) and \(6 - 4i\)
\((6 - 4i)+(6 - 4i)=12-8i\)
Step 2: Subtract \(3 - 2i\) from \(12-8i\)
\((12-8i)-(3 - 2i)=(12-3)+(-8i + 2i)=9-6i\)
Problem 8: Subtract \(8 - 6i\) from \(9 + 3i\)
Step 1: Rewrite the subtraction
\((9 + 3i)-(8 - 6i)=(9-8)+(3i + 6i)\)
Step 2: Combine real and imaginary parts
Real parts: \(9-8 = 1\)
Imaginary parts: \(3i+6i = 9i\)? Wait, no: \(3i-(-6i)=3i + 6i=9i\)? Wait, original subtraction: \((9 + 3i)+(-8 + 6i)\)
Real parts: \(9-8 = 1\)
Imaginary parts: \(3i+6i = 9i\)? Wait, no, the problem says "Subtract \(8 - 6i\) from \(9 + 3i\)", so it is \((9 + 3i)-(8 - 6i)=9 + 3i-8 + 6i=(9-8)+(3i + 6i)=1 + 9i\)? Wait, I made a mistake earlier. Let's correct:
\((9 + 3i)-(8 - 6i)=9-8+3i + 6i=1 + 9i\)
Problem 9: Add \(12 + 5i\), \(8 + 6i\) and \(8 - 6i\)
Step 1: Add the first two complex numbers
\((12 + 5i)+(8 + 6i)=20 + 11i\)
Step 2: Add the third complex number
\((20 + 11i)+(8 - 6i)=20 + 8+11i-6i=28 + 5i\)
Problem 10: Subtract \(15 - 6i\) from \(13 - 4i\)
Step 1: Rewrite the subtraction
\((13 - 4i)-(15 - 6i)=13 - 4i-15 + 6i\)
Step 2: Combine real and imaginary parts
Real parts: \(13-15=-2\)
Imaginary parts: \(-4i + 6i = 2i\)
Result: \(-2 + 2i\)
Problem 11: \((-2 + 3i)+(7 - 4i)-(4 + 5i)\)
Step 1: First, add the first two complex numbers
\((-2 + 3i)+(7 - 4i)=(-2 + 7)+(3i-4i)=5 - i\)
Step 2: Subtract the third complex number
\((5 - i)-(4 + 5i)=5 - i-4 - 5i=(5-4)+(-i-5i)=1-6i\)
Final Answers:
- \(\boldsymbol{3 + 5i}\)
- \(\boldsymbol{13-3i}\)
- \(\boldsymbol{3 + 3i}\)
- \(\boldsymbol{6-9i}\)
- \(\boldsymbol{18 + i}\)
- \(\boldsymbol{2-3i}\)
- \(\boldsymbol{9-6i}\) (corrected from earlier mistake in problem 8 analysis, problem 7 is as above)
- \(\boldsymbol{1 + 9i}\)
- \(\boldsymbol{28 + 5i}\)
- \(\boldsymbol{-2 + 2i}\)
- \(\boldsymbol{1-6i}\)