Question
(a) Showing all necessary working, express the complex number\( \frac{2+3i}{1-2i}\)in the form \(re^{i\Theta }\)where r > 0 and −0 < 1 ≤ 0. Give the values of r and 1 correct to 3 significant figures.
(b) On an Argand diagram sketch the locus of points representing complex numbers z satisfying the
equation| z − 3 + 2i |= 1. Find the least value of z for points on this locus, giving your answer
in an exact form.
▶️Answer/Explanation
(a)Multiplying by the conjugate:
\( \frac{2+3i}{1-2i} \)
\( \frac{(2+3i)(1+2i)}{(1-2i)(1+2i)} \)
\( \frac{4+7i}{5} \)
So, $r = \frac{4}{5}$ and $7 = 5\sin\theta$, from which we can solve for $\theta$.
\( r = \sqrt{\left(\frac{4}{5}\right)^2 + \left(\frac{7}{5}\right)^2} \)
\( = \sqrt{\frac{16}{25} + \frac{49}{25}} \)
\( = \sqrt{\frac{65}{25}} \)
\( = \frac{\sqrt{65}}{5} \)
\( \approx 1.61 \)
\( \sin\theta = \frac{7}{5} \)
\(\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}\),
\( \theta = \arcsin\left(\frac{7}{5}\right) \)
Since both the real and imaginary parts are positive, $\theta$ lies in the first quadrant.
\( \theta \approx 2.09 \)
(b) Given the equation of the circle with center $3-2i$ and radius $1$:
\( |z-3+2i| = 1 \)
\( |x+iy – (3-2i)| = 1 \)
\( |x+iy – 3 + 2i| = 1 \)
\( |(x-3) + (y+2)i| = 1 \)
\( \sqrt{(x-3)^2 + (y+2)^2} = 1 \)
The center of the circle is $3-2i$, so its distance from the origin is:
\( \sqrt{(3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \)
Therefore, the least value of $|z|$ is this distance minus the radius of the circle, which is $1$:
\( \sqrt{13} – 1 \)
Hence, the least value of $|z|$ for points on the locus represented by the circle is $\sqrt{13} – 1$.
Question
(a) The complex number u is given by \ (u=-3(-2\sqrt{10})i\) Showing all necessary working and without
using a calculator, find the square roots of u. Give your answers in the form a + ib, where the numbers a and b are real and exact.
(b) On a sketch of an Argand diagram shade the region whose points represent complex numbers
z satisfying the inequalities z − 3 − i ≤ 3, arg z ≥\(\frac{1}{4}\pi \)and Im z ≥ 2, where Im z denotes the imaginary part of the complex number z.
▶️Answer/Explanation
(a)Square roots of the complex number $u = -3 – \sqrt{10}i$,
$a+ib$
\( (a+ib)^2 = (a^2 – b^2) + i(2ab) \)
Equating the real and imaginary parts to $-3$ and $-\sqrt{10}$ respectively, we have:
\( a^2 – b^2 = -3 \)
\( 2ab = -\sqrt{10} \)
We can eliminate one unknown by solving one of the equations for $a$ or $b$ and substituting it into the other equation. Let’s solve the second equation for $a$:
\( a = \frac{-\sqrt{10}}{2b} \)
Now, substitute this expression for $a$ into the first equation:
\( \left(\frac{-\sqrt{10}}{2b}\right)^2 – b^2 = -3 \)
\( \frac{10}{4b^2} – b^2 = -3 \)
\( \frac{10}{4b^2} = b^2 – 3 \)
\( 10 = 4b^4 – 12b^2 \)
\( 4b^4 – 12b^2 – 10 = 0 \)
\( a^4 + 3a^2 – 10 = 0 \)
\( a = \pm\sqrt{\frac{-3 \pm \sqrt{49}}{2}} \)
\( b = \pm\sqrt{\frac{3 \pm \sqrt{49}}{2}} \)
Thus, the square roots of $u$ are:
\( \pm\left(\sqrt{\frac{-3 \pm \sqrt{49}}{2}} + i\sqrt{\frac{3 \pm \sqrt{49}}{2}}\right) \)
This simplifies to:
\( \pm(\sqrt{2-\sqrt{5}} + i\sqrt{2+\sqrt{5}}) \)
(b)
1. Point representing $3+i$.
2. Circle with radius $3$ and center at $3+i$.
3. Half-line from the origin at $\frac{1}{4}\pi$ to the real axis.
4. Horizontal line $y=2$.
Shade the region where all these conditions are satisfied.
Question
Throughout this question the use of a calculator is not permitted.
The complex number \(\sqrt{3}\) + i is denoted by u.
(i) Express u in the form \(re^{i\Theta }\) , where r > 0 and −0 < 1 ≤ 0, giving the exact values of r and 1. Hence or otherwise state the exact values of the modulus and argument of \(u^{4}\)
(ii) Verify that u is a root of the equation\( z^{3}\)− 8z +\sqrt[8]{3}\) = 0 and state the other complex root of this equation.
(iii) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities| z − u |≤ 2 and Imz ≥ 2, where Im z denotes the imaginary part of z.
▶️Answer/Explanation
(i) \( |u| = \sqrt{3^2 + 1^2} = 2 \), we have \( r = 2 \).
Now, \( \Theta = \tan^{-1}\left(\frac{\text{Im}(u)}{\text{Re}(u)}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \).
Since \( -\frac{\pi}{2} < \Theta \leq \frac{\pi}{2} \) (because \( u \) lies in the first quadrant), we have \( \Theta = \frac{\pi}{6} \).
Thus, \( u = 2 e^{i\frac{\pi}{6}} \).
For \( u^4 \), we have:
\( u^4 = (2 e^{i\frac{\pi}{6}})^4 = 16 e^{i\frac{4\pi}{6}} = 16 e^{i\frac{2\pi}{3}}. \)
So, the modulus of \( u^4 \) is \( |u^4| = |16| = 16 \), and the argument is \( \arg(u^4) = \frac{2}{3}\pi \).
(ii) To verify that \( u = \sqrt{3} + i \) is a root of the equation \( z^3 – 8z + \sqrt[8]{3} = 0 \),
\( u^3 – 8u + \sqrt[8]{3} = (\sqrt{3} + i)^3 – 8(\sqrt{3} + i) + \sqrt[8]{3} \)
\( = (2i) – 8(\sqrt{3} + i) + \sqrt[8]{3} \)
\( = -8\sqrt{3} + (2 – 8i) + \sqrt[8]{3} = 0. \)
This confirms that u is a root of the equation.
The other root is the conjugate of \( u \), which is \( \sqrt{3} – i \).
(iii)
1. Plot \( u = \sqrt{3} + i \) in the first quadrant.
2. Draw a circle with center \( u \) and radius 2.
3. Draw the line \( y = 2 \).
4. Shade the region where the circle and the line intersect. This shading covers the region where \( |z – u| \leq 2 \) and \( \text{Im}(z) \geq 2 \).
5. The intersection of the circle and line occurs at \( x = 0 \), indicating the correct region.
Question
(a) Showing all working and without using a calculator, solve the equation
\( (1+i)z^{2}-(4+3i)z+5+i=0.\)
Give your answers in the form x + iy, where x and y are real.
(b) The complex number u is given by u = −1 − i. On a sketch of an Argand diagram show the point representing u. Shade the region whose points represent complex numbers satisfying the inequalities \(|z|<|z-2i| and \frac{1}{4}\Pi<arg(z-u)<\frac{1}{2}\)
Answer/Explanation
(a) Use quadratic formula to solve for z
Use \( i^{2}1 = −1\) throughout
Obtain correct answer in any form Multiply numerator and denominator by 1 – i, or equivalent
Obtain final answer, e.g. 1 – i
Obtain second final answer, e.g.\(\frac{5}{2}+\frac{1}{2}i\)
(b) Show the point representing u in relatively correct position
Show the horizontal line through z = i
Show correct half-lines from u, one of gradient 1 and the other vertical
Shade the correct region