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Let $z_1 = 1 - 2i$, where $i^2 = -1$ - Leaving Cert Mathematics - Question 2 - 2014

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Let-$z_1-=-1---2i$,-where-$i^2-=--1$-Leaving Cert Mathematics-Question 2-2014.png

Let $z_1 = 1 - 2i$, where $i^2 = -1$. (a) The complex number $z_1$ is a root of the equation $2z^3 - 7z^2 + 16z - 15 = 0$. Find the other two roots of the equation.... show full transcript

Worked Solution & Example Answer:Let $z_1 = 1 - 2i$, where $i^2 = -1$ - Leaving Cert Mathematics - Question 2 - 2014

Step 1

Find the other roots of the equation

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Answer

Given the polynomial equation is:

2z37z2+16z15=02z^3 - 7z^2 + 16z - 15 = 0

Since z1=12iz_1 = 1 - 2i is a root, we can use polynomial division to factor out (z(12i))(z - (1 - 2i)) from the equation. The result of the division gives us:

(z(12i))(2z2+z15)(z - (1 - 2i))(2z^2 + z - 15)

Next, we can find the roots of the quadratic equation 2z2+z15=02z^2 + z - 15 = 0 using the quadratic formula:

z=b±b24ac2az = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Where a=2a = 2, b=1b = 1, and c=15c = -15.

Substituting these values, we calculate:

z=1±1+1204=1±1214=1±114z = \frac{-1 \pm \sqrt{1 + 120}}{4} = \frac{-1 \pm \sqrt{121}}{4} = \frac{-1 \pm 11}{4}

Thus, we obtain:

z=104=2.5andz=124=3z = \frac{10}{4} = 2.5 \\ \text{and} \\ z = \frac{-12}{4} = -3

Therefore, the other roots are z2=2.5z_2 = 2.5 and z3=3z_3 = -3.

Step 2

Plot $z$, $ar{z_1}$ and $w$ on the Argand diagram

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Answer

In the Argand diagram:

  • Plot the point for z=12iz = 1 - 2i, which is located at (1, -2).
  • The conjugate z1ˉ=1+2i\bar{z_1} = 1 + 2i is plotted at (1, 2).
  • For the point w=z1ˉz1w = \bar{z_1} z_1, first calculate: z1ˉz1=(1+2i)(12i)=12(2i)2=1+4=5\bar{z_1} z_1 = (1 + 2i)(1 - 2i) = 1^2 - (2i)^2 = 1 + 4 = 5 Plot this point at (5, 0).

Label each point accordingly on the diagram.

Step 3

Find the measure of the acute angle, $ar{z_1} w z_1$

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Answer

To find the angle formed by the points z1ˉ\bar{z_1}, ww, and z1z_1, we can use the tangent function:

The lengths are:

  • z1ˉw=(1+2i)5=(4+2i)=(4)2+(2)2=16+4=20=25|\bar{z_1} w| = |(1 + 2i) - 5| = |(-4 + 2i)| = \sqrt{(-4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}
  • z1w=(12i)5=(42i)=25|z_1 w| = |(1 - 2i) - 5| = |(-4 - 2i)| = 2\sqrt{5}

Using the tangent formula:

tan(θ)=oppositeadjacent=y2y1x2x1\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{|y_2 - y_1|}{|x_2 - x_1|}

Substituting the calculated values gives:

  • Opposite side: 2
  • Adjacent side: 4

Thus: tan1(24)=tan1(12)26.57\tan^{-1}(\frac{2}{4}) = \tan^{-1}(\frac{1}{2}) \approx 26.57^{\circ}

Final answer, rounding to the nearest degree:

θ53\theta \approx 53^{\circ}

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