Math worksheet with complex number problems.

Let's solve each of the problems step by step.

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1. Find the multiplicative inverse of $ 4 - 3i $.

The multiplicative inverse of a complex number $ z = a + bi $ is:

$$
\frac{1}{z} = \frac{\overline{z}}{|z|^2}
$$

For $ z = 4 - 3i $, conjugate $ \overline{z} = 4 + 3i $, and modulus squared:

$$
|z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25
$$

So,

$$
\frac{1}{4 - 3i} = \frac{4 + 3i}{25} = \frac{4}{25} + \frac{3}{25}i
$$

✔ Answer: $ \boxed{\frac{4}{25} + \frac{3}{25}i} $

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2. Express in terms of $ a + ib $:

$$
\frac{(3 + i\sqrt{5})(3 - i\sqrt{5})}{(\sqrt{3} + i\sqrt{2}) - (\sqrt{3} - i\sqrt{2})}
$$

Numerator:
$$
(3 + i\sqrt{5})(3 - i\sqrt{5}) = 3^2 - (i\sqrt{5})^2 = 9 - (-5) = 9 + 5 = 14
$$

Denominator:
$$
(\sqrt{3} + i\sqrt{2}) - (\sqrt{3} - i\sqrt{2}) = \sqrt{3} + i\sqrt{2} - \sqrt{3} + i\sqrt{2} = 2i\sqrt{2}
$$

So the expression becomes:
$$
\frac{14}{2i\sqrt{2}} = \frac{7}{i\sqrt{2}}
$$

Now simplify:
$$
\frac{7}{i\sqrt{2}} = \frac{7}{i\sqrt{2}} \cdot \frac{-i}{-i} = \frac{-7i}{\sqrt{2}} \cdot \frac{1}{i^2} = \frac{-7i}{\sqrt{2}} \cdot (-1) = \frac{7i}{\sqrt{2}}
$$

Rationalize:
$$
\frac{7i}{\sqrt{2}} = \frac{7i\sqrt{2}}{2}
$$

So, in the form $ a + ib $, we have:
$$
a = 0, \quad b = \frac{7\sqrt{2}}{2}
$$

✔ Answer: $ \boxed{0 + \frac{7\sqrt{2}}{2}i} $

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3. Evaluate $ i^n + i^{n+1} + i^{n+2} + i^{n+3} $

We know that powers of $ i $ cycle every 4:

- $ i^1 = i $
- $ i^2 = -1 $
- $ i^3 = -i $
- $ i^4 = 1 $
- $ i^5 = i $, etc.

So for any integer $ n $, consider:

$$
i^n + i^{n+1} + i^{n+2} + i^{n+3}
$$

Let’s factor $ i^n $:

$$
= i^n (1 + i + i^2 + i^3) = i^n (1 + i - 1 - i) = i^n (0) = 0
$$

✔ Answer: $ \boxed{0} $

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4. If $ 1, w, w^2 $ are cube roots of unity, show that $ (1 - w + w^2)(1 + w - w^2) = 4 $

Recall:
- $ 1 + w + w^2 = 0 $
- $ w^3 = 1 $
- $ w^2 = \overline{w} $, but more importantly, $ w^2 = -1 - w $

Let’s compute:

Let $ A = (1 - w + w^2)(1 + w - w^2) $

Use identity: $ (a + b)(a - b) = a^2 - b^2 $? Not quite symmetric.

But let’s expand directly:

Let $ A = (1 - w + w^2)(1 + w - w^2) $

Set $ a = 1 $, $ b = -w $, $ c = w^2 $, so first term: $ a + b + c $

Second: $ a + (-b) + (-c) $? Not helpful.

Just expand:

$$
A = 1 \cdot (1 + w - w^2) - w \cdot (1 + w - w^2) + w^2 \cdot (1 + w - w^2)
$$

Compute term by term:

1. $ 1(1 + w - w^2) = 1 + w - w^2 $

2. $ -w(1 + w - w^2) = -w - w^2 + w^3 = -w - w^2 + 1 $ (since $ w^3 = 1 $)

3. $ w^2(1 + w - w^2) = w^2 + w^3 - w^4 = w^2 + 1 - w $ (since $ w^4 = w $)

Now sum all three parts:

Add them:

First: $ 1 + w - w^2 $

Second: $ 1 - w - w^2 $

Third: $ 1 + w^2 - w $

Now add:

- Constants: $ 1 + 1 + 1 = 3 $
- $ w $ terms: $ w - w - w = -w $
- $ w^2 $ terms: $ -w^2 - w^2 + w^2 = -w^2 $

Wait — let's collect carefully:

From first: $ 1 + w - w^2 $

From second: $ 1 - w - w^2 $

From third: $ 1 - w + w^2 $

Sum:

Constants: $ 1 + 1 + 1 = 3 $

$ w $: $ w - w - w = -w $

$ w^2 $: $ -w^2 - w^2 + w^2 = -w^2 $

So total: $ 3 - w - w^2 $

But from $ 1 + w + w^2 = 0 $ → $ w + w^2 = -1 $

So $ -w - w^2 = 1 $

Thus: $ 3 + 1 = 4 $

✔ Answer: $ \boxed{4} $

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5. Find the sum product of the complex numbers $ -\sqrt{3} + \sqrt{2} $ and $ 2\sqrt{3} - i $

Wait: “sum product” is ambiguous. Probably means product of the two complex numbers.

Assume: find $ (-\sqrt{3} + \sqrt{2}) \cdot (2\sqrt{3} - i) $

Note: First number is real? Wait: $ -\sqrt{3} + \sqrt{2} $ is real, since no imaginary part.

So it's $ a = -\sqrt{3} + \sqrt{2} $, $ b = 2\sqrt{3} - i $

Then:

$$
(-\sqrt{3} + \sqrt{2})(2\sqrt{3} - i) = (-\sqrt{3})(2\sqrt{3}) + (-\sqrt{3})(-i) + \sqrt{2}(2\sqrt{3}) + \sqrt{2}(-i)
$$

Compute each term:

- $ -\sqrt{3} \cdot 2\sqrt{3} = -2 \cdot 3 = -6 $
- $ -\sqrt{3} \cdot (-i) = i\sqrt{3} $
- $ \sqrt{2} \cdot 2\sqrt{3} = 2\sqrt{6} $
- $ \sqrt{2} \cdot (-i) = -i\sqrt{2} $

Now combine:

Real part: $ -6 + 2\sqrt{6} $

Imaginary part: $ i\sqrt{3} - i\sqrt{2} = i(\sqrt{3} - \sqrt{2}) $

✔ Answer: $ \boxed{(-6 + 2\sqrt{6}) + i(\sqrt{3} - \sqrt{2})} $

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6. If $ a + ib = \frac{(x+i)^2}{2x - i} $, prove that $ a^2 + b^2 = \frac{(x^2 + 1)^2}{4x^2 + 1} $

We need to find $ |a + ib|^2 = a^2 + b^2 $

Since $ a + ib = \frac{(x+i)^2}{2x - i} $, then:

$$
|a + ib| = \left| \frac{(x+i)^2}{2x - i} \right| = \frac{|x+i|^2}{|2x - i|}
$$

So:

- $ |x + i| = \sqrt{x^2 + 1} $ → $ |x+i|^2 = x^2 + 1 $
- $ |2x - i| = \sqrt{(2x)^2 + (-1)^2} = \sqrt{4x^2 + 1} $

But $ |(x+i)^2| = |x+i|^2 = x^2 + 1 $

So:

$$
|a + ib| = \frac{|x+i|^2}{|2x - i|} = \frac{x^2 + 1}{\sqrt{4x^2 + 1}}
$$

Wait: actually:

$$
|a + ib| = \left| \frac{(x+i)^2}{2x - i} \right| = \frac{|x+i|^2}{|2x - i|} = \frac{x^2 + 1}{\sqrt{4x^2 + 1}}
$$

Then:

$$
a^2 + b^2 = |a + ib|^2 = \left( \frac{x^2 + 1}{\sqrt{4x^2 + 1}} \right)^2 = \frac{(x^2 + 1)^2}{4x^2 + 1}
$$

✔ Proved: $ \boxed{a^2 + b^2 = \frac{(x^2 + 1)^2}{4x^2 + 1}} $

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7. Evaluate $ \left[ i^{18} + \left( \frac{1}{i} \right)^{25} \right]^7 $

First, recall:

- $ i^1 = i $
- $ i^2 = -1 $
- $ i^3 = -i $
- $ i^4 = 1 $, cycle every 4

So:

- $ i^{18} = i^{4 \cdot 4 + 2} = i^2 = -1 $

Now $ \frac{1}{i} = -i $, because $ \frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i $

So $ \left( \frac{1}{i} \right)^{25} = (-i)^{25} $

Now $ (-i)^{25} = (-1)^{25} \cdot i^{25} = -1 \cdot i^{25} $

$ i^{25} = i^{4 \cdot 6 + 1} = i^1 = i $

So $ (-i)^{25} = -i $

Thus:

$$
i^{18} + \left( \frac{1}{i} \right)^{25} = -1 + (-i) = -1 - i
$$

Now raise to 7th power:

$$
(-1 - i)^7
$$

Write $ -1 - i $ in polar form.

Magnitude: $ |-1 - i| = \sqrt{1^2 + 1^2} = \sqrt{2} $

Argument: $ \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} $ (third quadrant)

So:

$$
-1 - i = \sqrt{2} \operatorname{cis} \left( \frac{5\pi}{4} \right)
$$

Raise to 7th power:

$$
(-1 - i)^7 = (\sqrt{2})^7 \operatorname{cis} \left( 7 \cdot \frac{5\pi}{4} \right) = 2^{7/2} \operatorname{cis} \left( \frac{35\pi}{4} \right)
$$

Simplify angle:

$ \frac{35\pi}{4} = 8\pi + \frac{3\pi}{4} = 4 \cdot 2\pi + \frac{3\pi}{4} $ → equivalent to $ \frac{3\pi}{4} $

So:

$$
= 2^{7/2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right) = 2^{7/2} \left( -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right)
$$

Now $ 2^{7/2} = 2^3 \cdot 2^{1/2} = 8\sqrt{2} $

So:

$$
8\sqrt{2} \cdot \left( -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = 8\sqrt{2} \cdot \frac{\sqrt{2}}{2} (-1 + i) = 8 \cdot \frac{2}{2} (-1 + i) = 8(-1 + i)
$$

Because $ \sqrt{2} \cdot \sqrt{2} = 2 $, and $ 8 \cdot \frac{2}{2} = 8 $

So:

$$
= -8 + 8i
$$

✔ Answer: $ \boxed{-8 + 8i} $

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8. Find modulus and argument of $ \frac{1 + i}{1 - i} $

Multiply numerator and denominator by conjugate of denominator:

$$
\frac{1 + i}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{(1 + i)^2}{(1)^2 - (i)^2} = \frac{1 + 2i + i^2}{1 - (-1)} = \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i
$$

So the expression simplifies to $ i $

Modulus: $ |i| = 1 $

Argument: $ \arg(i) = \frac{\pi}{2} $

✔ Answer: Modulus = $ \boxed{1} $, Argument = $ \boxed{\frac{\pi}{2}} $

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9. For what real values of $ x $ and $ y $ are the numbers equal: $ (1+i)y^2 $ and $ (2+i)x $?

We are given:

$$
(1+i)y^2 = (2+i)x
$$

Let’s write both sides as complex numbers.

Left: $ y^2 + i y^2 $

Right: $ 2x + i x $

Equate real and imaginary parts:

- Real: $ y^2 = 2x $
- Imaginary: $ y^2 = x $

Now set equal:

From above: $ y^2 = 2x $ and $ y^2 = x $

So $ 2x = x $ → $ x = 0 $

Then $ y^2 = x = 0 $ → $ y = 0 $

✔ Answer: $ \boxed{x = 0, \; y = 0} $

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10. Convert into polar form:

$$
z = \frac{i - 1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}
$$

First, note that denominator is $ \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \operatorname{cis} \left( \frac{\pi}{3} \right) $

Numerator: $ i - 1 = -1 + i $

So:

$$
z = \frac{-1 + i}{\operatorname{cis} \left( \frac{\pi}{3} \right)}
$$

We can write $ -1 + i $ in polar form.

Magnitude: $ \sqrt{(-1)^2 + 1^2} = \sqrt{2} $

Argument: $ \tan^{-1}(1 / -1) = \tan^{-1}(-1) $, but in second quadrant → $ \frac{3\pi}{4} $

So $ -1 + i = \sqrt{2} \operatorname{cis} \left( \frac{3\pi}{4} \right) $

Then:

$$
z = \frac{\sqrt{2} \operatorname{cis} \left( \frac{3\pi}{4} \right)}{\operatorname{cis} \left( \frac{\pi}{3} \right)} = \sqrt{2} \operatorname{cis} \left( \frac{3\pi}{4} - \frac{\pi}{3} \right)
$$

Compute angle:

$ \frac{3\pi}{4} - \frac{\pi}{3} = \frac{9\pi - 4\pi}{12} = \frac{5\pi}{12} $

So:

$$
z = \sqrt{2} \operatorname{cis} \left( \frac{5\pi}{12} \right)
$$

✔ Answer: Polar form: $ \boxed{\sqrt{2} \left( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right)} $

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## ✔ Final Answers Summary:

1. $ \boxed{\frac{4}{25} + \frac{3}{25}i} $
2. $ \boxed{0 + \frac{7\sqrt{2}}{2}i} $
3. $ \boxed{0} $
4. $ \boxed{4} $
5. $ \boxed{(-6 + 2\sqrt{6}) + i(\sqrt{3} - \sqrt{2})} $
6. Proved: $ \boxed{a^2 + b^2 = \frac{(x^2 + 1)^2}{4x^2 + 1}} $
7. $ \boxed{-8 + 8i} $
8. Modulus: $ \boxed{1} $, Argument: $ \boxed{\frac{\pi}{2}} $
9. $ \boxed{x = 0,\ y = 0} $
10. $ \boxed{\sqrt{2} \left( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right)} $