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 $-\frac{\pi}{4} \leq arg(z - 1) \leq \frac{\pi}{4}$ 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 Properties of moduli and arguments: 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.

96%

114 rated

Answer

To write $i^p$ in the form $a + ib$ , we first note that $i = e^{i\frac{\pi}{2}}$ . By using the exponential form, we can express this as:

i^p = \left(e^{i\frac{\pi}{2}}\right)^p = e^{i\frac{p\pi}{2}}.

Depending on the value of $p$ , this leads us to different points on the unit circle in the complex plane.

Step 2

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

99%

104 rated

Answer

The expression $-2 + 3i$ is already in the form $a + ib$ , where $a = -2$ and $b = 3$ .

Step 3

(i) the point $R$ representing $iz$

96%

101 rated

Answer

To find the point $R$ representing $iz$ , we multiply the complex number $z$ by $i$ . If $z = x + yi$ , then:

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

Thus, the coordinates for point $R$ will be $(-y, x)$ .

Step 4

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

98%

120 rated

Answer

The point $S$ representing $w$ is directly marked based on the coordinates of the complex number $w = a + bi$ . Therefore, the point $S$ is at $(a, b)$ on the Argand diagram.

Step 5

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

97%

117 rated

Answer

The point $T$ representing $z + w$ is calculated by adding the complex numbers:

T = (x + a) + (y + b)i.

This indicates that point $T$ is located at $(x + a, y + b)$ on the Argand 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.

97%

121 rated

Answer

The inequality $|z - 1| \leq 2$ represents a circle centered at $(1, 0)$ with a radius of 2. The argument condition limits the region to a sector, which can be visualized as a wedge extending from the center of the circle at $z = 1$ , within the angles of $-\frac{\pi}{4}$ and $\frac{\pi}{4}$ . Therefore, the intersection of these two regions will be drawn accordingly.

Step 7

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

96%

114 rated

Answer

Using De Moivre's theorem, we can express $-1$ in polar form as:

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

To find the 5th roots:

Calculate the modulus: $|z| = 1$ .
Find the argument: $arg(z) = \frac{\pi + 2k\pi}{5}$ for $k = 0, 1, 2, 3, 4$ .

So, the 5th roots of $-1$ are:

z_k = 1^{\frac{1}{5}} \cdot e^{i\left(\frac{\pi + 2k\pi}{5}\right)}, \quad k = 0, 1, 2, 3, 4.

Step 8

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

99%

104 rated

Answer

The 5th roots of $-1$ will be plotted as five equally spaced points around the unit circle, each separated by an angle of $\frac{2\pi}{5}$ . Specifically, they will lie along the unit circle at angles:

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

Step 9

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

96%

101 rated

Answer

To find the square roots of $3 + 4i$ , express it in polar form:

Calculate the modulus: $r = \sqrt{3^2 + 4^2} = 5$ .
Calculate the argument: $\theta = \tan^{-1}\left(\frac{4}{3}\right)$ .

Now we can write:

3 + 4i = 5 \cdot e^{i\theta}.

To find the square roots, we use:

z = \sqrt{5} \cdot e^{i\frac{\theta}{2}}.

This will yield two solutions based on the periodic nature of the angles.

Step 10

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

98%

120 rated

Answer

Rearranging gives:

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

Using the quadratic formula:

z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\quad a=1, b=i, c=i-1.

Calculate the discriminant and simplify to find 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.

(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.

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

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

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

Join the SSCE students using SimpleStudy...