Write $i^p$ in the form $a + ib$ where $a$ and $b$ are real. Write $-2 + 3i$ in the form $a + ib$ where $a$ and $b$ are real. The points $P$ and $Q$ on the Argand diagram represent the complex numbers $z$ and $w$ respectively. Copy the diagram into your writing booklet, and mark on it the following points: (i) the point $R$ representing $iz$ (ii) the point $S$ representing $w$ (iii) the point $T$ representing $z + w$. Sketch the region in the complex plane where the inequalities $|z - 1| \,|\leq 2$ and $-rac{\pi}{4} \,|\leq \arg(z - 1) \leq \frac{\pi}{4}$ hold simultaneously. Find all the 5th roots of $-1$ in modulus-argument form. Sketch the 5th roots of $-1$ on an Argand diagram. Find the square roots of $3 + 4i$. Hence, or otherwise, solve the equation $z^2 + iz - 1 = -i$.

SSCE HSC Mathematics Extension 2 Complex numbers: Write $i^p$ in the form $a + ib$

Write $i^p$ in the form $a + ib$ where $a$ and $b$ are real - HSC - SSCE Mathematics Extension 2 - Question 2 - 2009 - Paper 1

Question 2

Write $i^p$ in the form $a + ib$ where $a$ and $b$ are real. Write $-2 + 3i$ in the form $a + ib$ where $a$ and $b$ are real. The points $P$ and $Q$ on the Argand ... show full transcript

Worked Solution & Example Answer:Write $i^p$ in the form $a + ib$ where $a$ and $b$ are real - HSC - SSCE Mathematics Extension 2 - Question 2 - 2009 - Paper 1

Step 1

Write $i^p$ in the form $a + ib$ where $a$ and $b$ are real.

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Answer

To express $i^p$ in the form $a + ib$ , we use Euler's formula. Recall that $i = e^{i\frac{\pi}{2}}$ . Thus, we have:

$i^p = (e^{i\frac{\pi}{2}})^p = e^{i\frac{p\pi}{2}} = \cos\left(\frac{p\pi}{2}\right) + i\sin\left(\frac{p\pi}{2}\right)$

This represents $a = \cos\left(\frac{p\pi}{2}\right)$ and $b = \sin\left(\frac{p\pi}{2}\right)$ .

Step 2

Write $-2 + 3i$ in the form $a + ib$ where $a$ and $b$ are real.

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Answer

The expression $-2 + 3i$ is already in the desired form where $a = -2$ and $b = 3$ . Therefore, no conversion is needed.

Step 3

Copy the diagram into your writing booklet, and mark on it the following points: (i) the point $R$ representing $iz$

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Answer

To find point $R$ representing $iz$ , we multiply the coordinates of $z$ by $i$ . Given that $z = x + yi$ , we have:

$R = i(x + yi) = -y + xi$

Mark this point on the diagram accordingly.

Step 4

(ii) the point $S$ representing $w$

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Answer

Mark the point $S$ on the Argand diagram directly corresponding to the complex number $w$ . This is the Cartesian coordinates $(\text{Re}(w), \text{Im}(w))$ .

Step 5

(iii) the point $T$ representing $z + w$.

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Answer

To find point $T$ that represents $z + w$ , we add the real and imaginary parts of $z$ and $w$ :

$T = (x_1 + x_2) + i(y_1 + y_2)$

Mark this resultant point on the diagram.

Step 6

Sketch the region in the complex plane where the inequalities $|z - 1| \leq 2$ and $-\frac{\pi}{4} \leq \arg(z - 1) \leq \frac{\pi}{4}$ hold simultaneously.

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The inequality $|z - 1| \leq 2$ describes a circle centered at $1$ with radius $2$ . The second inequality, $-\frac{\pi}{4} \leq \arg(z - 1) \leq \frac{\pi}{4}$ , describes a sector of the circle. Draw this circle and highlight the corresponding sector that meets these conditions.

Step 7

Find all the 5th roots of $-1$ in modulus-argument form.

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Answer

First, express $-1$ in polar form:

$-1 = 1 \cdot e^{i\pi}$

The formula for the $n$ th roots is:

$z_k = r^{1/n} e^{i(\theta + 2k\pi)/n}$

For $k = 0, 1, 2, 3, 4$ (5 roots), we find:

$z_k = 1^{1/5} e^{i(\pi + 2k\pi)/5}$

Thus, the 5th roots are:

$z_k = e^{i\frac{\pi + 2k\pi}{5}}$

Step 8

(ii) Sketch the 5th roots of $-1$ on an Argand diagram.

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Answer

Plot the points corresponding to the calculated 5th roots on the Argand diagram. They will be evenly spaced around the unit circle at angles of:

$\frac{\pi}{5}, \frac{3\pi}{5}, \frac{\pi}{5}, \frac{7\pi}{5}, \frac{9\pi}{5}$

Step 9

Find the square roots of $3 + 4i$.

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Answer

To find the square roots of a complex number, we can set:

$z = x + yi$

Then, we square it:

$(x + yi)^2 = 3 + 4i \Rightarrow x^2 - y^2 + 2xyi = 3 + 4i$

This gives us two equations:

$x^2 - y^2 = 3$
$2xy = 4 \Rightarrow xy = 2$

We can solve this system simultaneously for $x$ and $y$ .

Step 10

Hence, or otherwise, solve the equation $z^2 + iz - 1 = -i$.

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Answer

Rearranging the equation gives:

$z^2 + iz + i - 1 = 0$

This is a quadratic equation in standard form. We apply the quadratic formula:

$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$ , $b = i$ , and $c = i - 1$ . Calculate the discriminant and then solve for the roots.

Write $i^p$ in the form $a + ib$ where $a$ and $b$ are real - HSC - SSCE Mathematics Extension 2 - Question 2 - 2009 - Paper 1

Worked Solution & Example Answer:Write $i^p$ in the form $a + ib$ where $a$ and $b$ are real - HSC - SSCE Mathematics Extension 2 - Question 2 - 2009 - Paper 1

Write $i^p$ in the form $a + ib$ where $a$ and $b$ are real.

Write $-2 + 3i$ in the form $a + ib$ where $a$ and $b$ are real.

Copy the diagram into your writing booklet, and mark on it the following points: (i) the point $R$ representing $iz$

(ii) the point $S$ representing $w$

(iii) the point $T$ representing $z + w$.

Sketch the region in the complex plane where the inequalities $|z - 1| \leq 2$ and $-\frac{\pi}{4} \leq \arg(z - 1) \leq \frac{\pi}{4}$ hold simultaneously.

Find all the 5th roots of $-1$ in modulus-argument form.

(ii) Sketch the 5th roots of $-1$ on an Argand diagram.

Find the square roots of $3 + 4i$.

Hence, or otherwise, solve the equation $z^2 + iz - 1 = -i$.

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