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2. \( z_1 = -3 + 4i \) and \( z_2 = 4 + 3i \), where \( i^2 = -1 \) - Leaving Cert Mathematics - Question 2 - 2021
Question 2
2. \( z_1 = -3 + 4i \) and \( z_2 = 4 + 3i \), where \( i^2 = -1 \). (a) Plot and label \( z_1 \), \( z_2 \), and \( z_1 + z_2 \) on the Argand Diagram. (b) \(... show full transcript
Worked Solution & Example Answer:2. \( z_1 = -3 + 4i \) and \( z_2 = 4 + 3i \), where \( i^2 = -1 \) - Leaving Cert Mathematics - Question 2 - 2021
Step 1
Plot and label \( z_1 \), \( z_2 \), and \( z_1 + z_2 \) on the Argand Diagram.
Answer
To plot the complex numbers, we break them down as follows:
-
Locate ( z_1 ): ( z_1 = -3 + 4i ) corresponds to the point (-3, 4) on the Argand diagram.
-
Locate ( z_2 ): ( z_2 = 4 + 3i ) corresponds to the point (4, 3).
-
Calculate ( z_1 + z_2 ):
[ z_1 + z_2 = (-3 + 4i) + (4 + 3i) = 1 + 7i ]
The coordinates for ( z_1 + z_2 ) are (1, 7). -
Plot all three points on the Argand plane and label them appropriately.
Step 2
Find \( z_3 = \frac{z_1}{z_2} \) in the form \( a + bi \).
Answer
Starting with the complex division:
[
z_3 = \frac{z_1}{z_2} = \frac{-3 + 4i}{4 + 3i}
]
To simplify, we multiply the numerator and the denominator by the conjugate of the denominator:
[ z_3 = \frac{(-3 + 4i)(4 - 3i)}{(4 + 3i)(4 - 3i)} ]
Evaluating the denominator:
[ (4 + 3i)(4 - 3i) = 16 + 9 = 25 ]
Evaluate the numerator:
[ (-3 + 4i)(4 - 3i) = -12 + 9i + 16i + 12 = 0 + 25i ] Thus:
[
z_3 = \frac{25i}{25} = i = 0 + 1i
]
Therefore, ( a = 0 ) and ( b = 1 ).
Step 3
Find \( |z_1 - \bar{z_1}| \) where \( \bar{z_1} \) is the complex conjugate of \( z_1 \).
Answer
The complex conjugate ( \bar{z_1} ) is given by changing the sign of the imaginary part:
[
\bar{z_1} = -3 - 4i
]
Now we find ( z_1 - \bar{z_1} ):
[
z_1 - \bar{z_1} = (-3 + 4i) - (-3 - 4i) = 8i
]
We then find the modulus:
[
|z_1 - \bar{z_1}| = |8i| = 8
]
There is no need for the form ( p\sqrt{q} ) since ( 8 = 8\sqrt{1} ), so ( p = 8 ) and ( q = 1 ).