Practise Questions Flashcards
i5 = ?
i
i7 = ?
-i
Solve x2 + 9 = 0
x2 - 32i2 = 0
(x + 3i)(x - 3i) = 0
x = ±3i
Solve -2x2 + 2x - 13 = 0
2x2 - 2x + 13 = 0
x = 2 - √<span>4 - 104</span>⁄4
x = 2 - √<span>100</span>⁄4
x = 2 - 10i⁄4
x = 1 - 5i⁄2
x = 1⁄2 - 5⁄2i
Simplify (4 + 7i) - (-2 + 9i)
4 + 7i + 2 - 9i = 6 - 2i
Simplify (5 + 2i)(3 - 4i)
= 15 - 20i + 6i - 8i2
= 15 - 14i + 8
= 23 - 14i
If z = 5 - 2i
z = ?
5 + 2i
If z1 = 3 + i and z2 = 2 - 3i, find (z1 - z2)2
= (z1 - z2)2
= (3 + i - 2 + 3i)2
= (1 + 4i)2
= 1 + 8i + 16i2
= 1 + 8i - 16
= -15 + 8i
Find real numbers x and y if (1 + i)x + (2 - 3i)y = 10
(1 + i)x + (2 - 3i)y = 10
- x* + ix + 2y - 3iy = 10
(1) x + 2y = 10
(2) x - 3y = 0
5y = 10 Subtract (2) from (1)
- y* = 2
- x* = 6
Solve 2z - 1 = (4 - i)2
2x + 2iy - 1 = 16 - 8i + i2
2x + 2iy - 1 = 16 - 8i - 1
2x + 2iy - 1 = 15 - 8i
2x + 2iy - 16 + 8i = 0
- x* - 8 + iy + 4i = 0
- x* = 8
- y* = -4
Factorise z2 + 16
z2 - 16i2 = (z + 4i)(z - 4i)
(z = ±4i)
Solve z2 + 2z + 26 = 0
- z* = -2 ±√<span>4 - 104</span>⁄2
- z* = -2 ±10i⁄2
- z* = -1 ± 10i
Find the square root of 12 + 5i
CURRENTLY WRONG REVISE
(x + iy)2 = 12 + 5i
- x2 - y2 + 2ixy* = 12 + 5i
(1) x2 - y2 = 12
(2) 2xy = 5 - x* = 5⁄2<em>y</em>
25⁄4<em>y</em>2 - <em>y</em>2 = 12
25 - 4y4 = 48y2
4y4 + 48y2 - 25 = 0
- y*2 = -48 ±√2304 + 400⁄2
- y*2 = 2 OR -50
- y* = ±√2 (y is an element of the real)
- x* = ±5⁄2√<span>2</span>
- z* = ±(5⁄2√<span>2</span> + √2i)
Solve z2 - (3 - 2i)z + (1 - 3i) = 0
z2 - (3 - 2i)z + (1 - 3i) = 0
(x2 - y2- 3x - 2y + 1) + i(2xy - 3y + 2x - 3) = 0
(1) x2 - y2- 3x - 2y + 1 = 0
(2) 2xy - 3y + 2x - 3 = 0