Find the two complex numbers z₁ and z₂ that satisfy the following simultaneous equations, where i² = −1: iz₁ = −4 + 3i 3z₁ − z₂ = 11 + 17i. Write your answers in the form a + bi where a, b ∈ ℤ. The complex numbers 3 + 2i and 5 − i are the first two terms of a geometric sequence. (i) Find r, the common ratio of the sequence. Write your answer in the form a + bi where a, b ∈ ℤ. (ii) Use de Moivre's Theorem to find T₉, the ninth term of the sequence. Write your answer in the form a + bi where a, b ∈ ℤ.

Leaving Cert Mathematics Complex Numbers: Find the two complex numbers z₁

Find the two complex numbers z₁ and z₂ that satisfy the following simultaneous equations, where i² = −1: iz₁ = −4 + 3i 3z₁ − z₂ = 11 + 17i - Leaving Cert Mathematics - Question 2 - 2020

Question 2

Find-the-two-complex-numbers-z₁-and-z₂-that-satisfy-the-following-simultaneous-equations,-where-i²-=-−1:--iz₁-=-−4-+-3i-3z₁-−-z₂-=-11-+-17i-Leaving Cert Mathematics-Question 2-2020.png

Find the two complex numbers z₁ and z₂ that satisfy the following simultaneous equations, where i² = −1: iz₁ = −4 + 3i 3z₁ − z₂ = 11 + 17i. Write your answers in t... show full transcript

Worked Solution & Example Answer:Find the two complex numbers z₁ and z₂ that satisfy the following simultaneous equations, where i² = −1: iz₁ = −4 + 3i 3z₁ − z₂ = 11 + 17i - Leaving Cert Mathematics - Question 2 - 2020

Step 1

Find z₁ and z₂ that satisfy the equations

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Answer

To solve the equations, start with the first equation:

$i z_1 = -4 + 3i$ Substituting $i$ :

$z_1 = \frac{-4 + 3i}{i} = -4i - 3 = -3 - 4i$

Now substituting into the second equation:

$3z_1 - z_2 = 11 + 17i$

Substituting for $z_1$ :

$3(-3 - 4i) - z_2 = 11 + 17i$

Which simplifies to:

$-9 - 12i - z_2 = 11 + 17i$

Rearranging gives:

$z_2 = -20 - 29i$

Thus, the two complex numbers are: $z_1 = -3 - 4i$ $z_2 = -20 - 29i$

Step 2

Find r, the common ratio of the sequence

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Answer

Given the first two terms:

1st term: $T_1 = 3 + 2i$
2nd term: $T_2 = 5 - i$

The common ratio $r$ is defined as:

$r = \frac{T_2}{T_1} = \frac{5 - i}{3 + 2i}$

To compute this, multiply numerator and denominator by the conjugate of the denominator:

$r = \frac{(5 - i)(3 - 2i)}{(3 + 2i)(3 - 2i)} = \frac{15 - 10i - 3i + 2}{9 + 4} = \frac{17 - 13i}{13}$

Thus, the common ratio is: $r = 1 - \frac{13}{13}i = 1 - i$

Step 3

Use de Moivre's Theorem to find T₉

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Answer

Using de Moivre's Theorem, we express the nth term:

$T_n = T_1 r^{n-1}$

Here, for $n = 9$ :

$T_9 = (3 + 2i)(1 - i)^{8}$

To calculate $(1 - i)^{8}$ , first find modulus and argument:

Modulus: $|1 - i| = \sqrt{2}$
Argument: $\tan^{-1}\left(-1\right) = -\frac{\pi}{4}$

Thus:

$(1 - i)^{8} = (\sqrt{2})^{8} \left(\cos(-2\pi) + i \sin(-2\pi)\right) = 16 (1 + 0i) = 16$

Therefore: $T_9 = (3 + 2i)(16) = 48 + 32i$

Find the two complex numbers z₁ and z₂ that satisfy the following simultaneous equations, where i² = −1: iz₁ = −4 + 3i 3z₁ − z₂ = 11 + 17i - Leaving Cert Mathematics - Question 2 - 2020

Worked Solution & Example Answer:Find the two complex numbers z₁ and z₂ that satisfy the following simultaneous equations, where i² = −1: iz₁ = −4 + 3i 3z₁ − z₂ = 11 + 17i - Leaving Cert Mathematics - Question 2 - 2020

Find z₁ and z₂ that satisfy the equations

Find r, the common ratio of the sequence

Use de Moivre's Theorem to find T₉

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