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Math  /  Algebra

Question14. Given the following complex numbers find z1+z2,z1⋅z2z_{1}+z_{2}, z_{1} \cdot z_{2} and z1z2\frac{z_{1}}{z_{2}} : z1=−3−4iz2=−23+2i\begin{array}{l} z_{1}=-3-4 i \\ z_{2}=-2 \sqrt{3}+2 i \end{array}

Studdy Solution

STEP 1

What is this asking? We're given two complex numbers, z1z_1 and z2z_2, and we need to find their sum, their product, and what happens when we divide z1z_1 by z2z_2. Watch out! Remember that i⋅i=−1i \cdot i = -1, and be careful with those negative signs when distributing!
Also, remember how to rationalize the denominator of a complex number by multiplying by its conjugate.

STEP 2

1. Calculate z1+z2z_1 + z_2
2. Calculate z1⋅z2z_1 \cdot z_2
3. Calculate z1z2\frac{z_1}{z_2}

STEP 3

Let's **add** the **real** parts and the **imaginary** parts separately!
We have z1=−3−4iz_1 = -3 - 4i and z2=−23+2iz_2 = -2\sqrt{3} + 2i.

STEP 4

Adding the real parts gives us −3+(−23)=−3−23-3 + (-2\sqrt{3}) = -3 - 2\sqrt{3}.

STEP 5

Adding the imaginary parts gives us −4i+2i=(−4+2)i=−2i-4i + 2i = (-4 + 2)i = -2i.

STEP 6

So, z1+z2=(−3−23)+(−2i)=−3−23−2iz_1 + z_2 = (-3 - 2\sqrt{3}) + (-2i) = -3 - 2\sqrt{3} - 2i.

STEP 7

Now, let's **multiply** z1z_1 and z2z_2 using the distributive property, just like we would with any two binomials!
Remember that i⋅i=−1i \cdot i = -1.

STEP 8

We have (−3−4i)(−23+2i)=(−3)(−23)+(−3)(2i)+(−4i)(−23)+(−4i)(2i)(-3 - 4i)(-2\sqrt{3} + 2i) = (-3)(-2\sqrt{3}) + (-3)(2i) + (-4i)(-2\sqrt{3}) + (-4i)(2i).

STEP 9

Simplifying each term gives us 63−6i+83i−8i26\sqrt{3} - 6i + 8\sqrt{3}i - 8i^2.

STEP 10

Since i2=−1i^2 = -1, we have 63−6i+83i−8(−1)=63−6i+83i+86\sqrt{3} - 6i + 8\sqrt{3}i - 8(-1) = 6\sqrt{3} - 6i + 8\sqrt{3}i + 8.

STEP 11

Combining like terms, we get (8+63)+(83−6)i(8 + 6\sqrt{3}) + (8\sqrt{3} - 6)i.
So, z1⋅z2=8+63+(83−6)iz_1 \cdot z_2 = 8 + 6\sqrt{3} + (8\sqrt{3} - 6)i.

STEP 12

To **divide** z1z_1 by z2z_2, we'll multiply both the numerator and denominator by the **conjugate** of the denominator.
The conjugate of −23+2i-2\sqrt{3} + 2i is −23−2i-2\sqrt{3} - 2i.

STEP 13

So, we have −3−4i−23+2i⋅−23−2i−23−2i\frac{-3 - 4i}{-2\sqrt{3} + 2i} \cdot \frac{-2\sqrt{3} - 2i}{-2\sqrt{3} - 2i}.

STEP 14

Multiplying the numerators gives us (−3−4i)(−23−2i)=(−3)(−23)+(−3)(−2i)+(−4i)(−23)+(−4i)(−2i)=63+6i+83i+8i2=63+6i+83i−8=63−8+(6+83)i(-3 - 4i)(-2\sqrt{3} - 2i) = (-3)(-2\sqrt{3}) + (-3)(-2i) + (-4i)(-2\sqrt{3}) + (-4i)(-2i) = 6\sqrt{3} + 6i + 8\sqrt{3}i + 8i^2 = 6\sqrt{3} + 6i + 8\sqrt{3}i - 8 = 6\sqrt{3} - 8 + (6 + 8\sqrt{3})i.

STEP 15

Multiplying the denominators gives us (−23+2i)(−23−2i)=(−23)2−(2i)2=12−4i2=12−4(−1)=12+4=16(-2\sqrt{3} + 2i)(-2\sqrt{3} - 2i) = (-2\sqrt{3})^2 - (2i)^2 = 12 - 4i^2 = 12 - 4(-1) = 12 + 4 = 16.

STEP 16

Therefore, z1z2=63−8+(6+83)i16=63−816+6+8316i=33−48+3+438i\frac{z_1}{z_2} = \frac{6\sqrt{3} - 8 + (6 + 8\sqrt{3})i}{16} = \frac{6\sqrt{3} - 8}{16} + \frac{6 + 8\sqrt{3}}{16}i = \frac{3\sqrt{3} - 4}{8} + \frac{3 + 4\sqrt{3}}{8}i.

STEP 17

We found that z1+z2=−3−23−2iz_1 + z_2 = -3 - 2\sqrt{3} - 2i, z1⋅z2=8+63+(83−6)iz_1 \cdot z_2 = 8 + 6\sqrt{3} + (8\sqrt{3} - 6)i, and z1z2=33−48+3+438i\frac{z_1}{z_2} = \frac{3\sqrt{3} - 4}{8} + \frac{3 + 4\sqrt{3}}{8}i.

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