Leaving Cert Mathematics Complex Numbers: The complex numbers $z_1$, $z

The complex numbers $z_1$, $z_2$ and $z_3$ are such that $$rac{2}{z_1} - rac{1}{z_2} = z_2 = 2 + 3i \text{ and } z_3 = 3 - 2i,$$ where $i^2 = -1$ - Leaving Cert Mathematics - Question 4 - 2015

Question 4

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The complex numbers $z_1$, $z_2$ and $z_3$ are such that $$rac{2}{z_1} - rac{1}{z_2} = z_2 = 2 + 3i \text{ and } z_3 = 3 - 2i,$$ where $i^2 = -1$. Write $z_1$ in... show full transcript

Worked Solution & Example Answer:The complex numbers $z_1$, $z_2$ and $z_3$ are such that $$rac{2}{z_1} - rac{1}{z_2} = z_2 = 2 + 3i \text{ and } z_3 = 3 - 2i,$$ where $i^2 = -1$ - Leaving Cert Mathematics - Question 4 - 2015

Step 1

The complex numbers $z_1$, $z_2$ and $z_3$ are such that

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Answer

Given that:

$z_2 = 2 + 3i$
$z_3 = 3 - 2i$

We start with the equation:

$\frac{2}{z_1} - \frac{1}{z_2} = z_2$

Substituting $z_2$ :

$\frac{2}{z_1} - \frac{1}{2 + 3i} = 2 + 3i$

Next, we need to find the conjugate of the denominator for the second term:

$\frac{1}{2 + 3i} = \frac{2 - 3i}{(2 + 3i)(2 - 3i)} = \frac{2 - 3i}{13}$

So, the equation becomes:

$\frac{2}{z_1} - \frac{2 - 3i}{13} = 2 + 3i$

Rearranging gives:

$\frac{2}{z_1} = 2 + 3i + \frac{2 - 3i}{13}$

Finding a common denominator:

$\frac{2}{z_1} = \frac{26 + 39i + 2 - 3i}{13} = \frac{28 + 36i}{13}$

Thus, we have:

$z_1 = \frac{2 \cdot 13}{28 + 36i} = \frac{26}{28 + 36i}$

Multiplying the numerator and denominator by the conjugate of the denominator:

$z_1 = \frac{26(28 - 36i)}{28^2 + 36^2} = \frac{728 - 936i}{1600} = \frac{91 - 117i}{200}$

So, in the form $a + bi$ , we find:

$z_1 = \frac{91}{200} - \frac{117}{200}i$

Step 2

Let $\omega$ be a complex number such that $\omega^3 = 1$, $\omega \neq 1$, and $S = 1 + \omega + \omega^2 + \ldots + \omega^{n-1}$.

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Answer

We can use the formula for the sum of a finite geometric series:

$S = \frac{a(1 - r^n)}{1 - r}$

where $a = 1$ , $r = \omega$ , and $n$ is the number of terms.

Thus,

$S = \frac{1(1 - \omega^n)}{1 - \omega}$

But since $\omega^3 = 1$ , we observe:

If $n$ is a multiple of 3, then $\omega^n = 1$ , leading to:

$S = \frac{0}{1 - \omega} = 0$

If $n$ is not a multiple of 3, then we calculate $S = \frac{1 - \omega^n}{1 - \omega}.$

This gives the value of $S$ depending on the divisibility of $n$ by 3.

The complex numbers $z_1$, $z_2$ and $z_3$ are such that $$ rac{2}{z_1} - rac{1}{z_2} = z_2 = 2 + 3i \text{ and } z_3 = 3 - 2i,$$ where $i^2 = -1$ - Leaving Cert Mathematics - Question 4 - 2015

Worked Solution & Example Answer:The complex numbers $z_1$, $z_2$ and $z_3$ are such that $$ rac{2}{z_1} - rac{1}{z_2} = z_2 = 2 + 3i \text{ and } z_3 = 3 - 2i,$$ where $i^2 = -1$ - Leaving Cert Mathematics - Question 4 - 2015

The complex numbers $z_1$, $z_2$ and $z_3$ are such that

Let $\omega$ be a complex number such that $\omega^3 = 1$, $\omega \neq 1$, and $S = 1 + \omega + \omega^2 + \ldots + \omega^{n-1}$.

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The complex numbers $z_1$, $z_2$ and $z_3$ are such that $$rac{2}{z_1} - rac{1}{z_2} = z_2 = 2 + 3i \text{ and } z_3 = 3 - 2i,$$ where $i^2 = -1$ - Leaving Cert Mathematics - Question 4 - 2015

Worked Solution & Example Answer:The complex numbers $z_1$, $z_2$ and $z_3$ are such that $$rac{2}{z_1} - rac{1}{z_2} = z_2 = 2 + 3i \text{ and } z_3 = 3 - 2i,$$ where $i^2 = -1$ - Leaving Cert Mathematics - Question 4 - 2015