Let the given complex number
z = x + ix =
We have to find the standard form of complex number.
Solution:
∴ x + iy =
Rationalising numerator part of complex number, we get
x + iy =
⇒ x + iy =
Using the algebraic identity:
(a + b)(a - b) = -
⇒ x + iy =
⇒ x + iy = [ ∵
]
⇒ x + iy =
⇒ x + iy =
⇒ x + iy =
⇒ x + iy = 1 - i
Thus, the given complex number in standard form as "1 - i".
Answer:
Step-by-step explanation:
slopes and equations:
find the equation thur ( 6,1 ) and (-2,-3)
find the slope m
m = (y2-y1) / (x2-x1 )
m = (-3 - 1) / (-2 -6)
m = -4 / -8
m = 1/2
now use the point-slope formula with our known slope
y-y1 = m(x-x1)
y-1 = 1/2(x-6)
y - 1 = 1/2x -3
y = 1/2x -3 +1
y = x -2
Find the equation parallel to y = 3x + 6 and thur (0,1)
Parallel means the same slope, the slope is 3 for the equation above.
use the slope-intercept formula again with the point given and the slope 3
y-1 = 3(x -0)
y - 1 = 3x
y = 3x +1
Find the equations perpendicular to 2x + y = 8 and the same y intercept as 4y = x + 3.
put both equations into proper form
y = -2x +6
y = x +
perpendicular means reciprocal slope and change the sign, the 2nd equation has an intercept of , so
y = 1/2x +
there you go Amanda :)
Parameterize the circular part of (call it
) by
wih , and the linear part (call it
) by
with .
Then