I have to do these without a calculator so explaining how you worked out the fractions would be very helpful, thank you :)

Question 1

4z + z(with the bar ontop) = 10 - 3i

what does z=

Question 2

z = i and w = -2 + 2i

arg(z) = ? π

arg(w) = ? π

Question 3

{ z € ℂ :  l z - i l ≤ 1 ]

what would the graph of this look like with a circle on it?

Question 4

y' = 4 / x^(1/3)

y = ? x^(2/3) + ?

when x and y = 1

Question 5

∫ 4x^(5/3) dx = ?

and

y = x^3, what is the area between x = 0 and x = 3?

Question 6

∫ (1 + x)dx   [with 2 at the top of the definite integral and 1 at the bottom] not sure how to put it in]

Question 7

What is the tangent to y = 3x^1/3 at (1,3)

AND

y = 7x^2 - 14x

What value is the minimum?

y'  = 14x - 14 (x = 1)

y'' = 14  and then times 1 i put 14, could you please tell me what i did wrong

asked Oct 8, 2014 in CALCULUS

1)

Let z = x + iy

4z + z(with the bar ontop) = 10 - 3i

4 (x +iy) +(x-iy) = 10 - 3i

4x+4iy + x  - iy  = 10 - 3i

5x + 3iy = 10 - 3i

Compare real and imaginary parts

5x = 10

x =2

3y = -3

y = -1

z = x+iy = 2 - 1i

selected Oct 10, 2014 by zoe

Question 2)

The complex number z = i

It can be written as 0 + i

Compare it to complex number  x + i y

x = 0, y = 1

In this case x = 0, y > 0

Since x = 0, y > 0 use the formula b = tan-1(y/x)

θ = tan-1 (1/0)

θ = tan-1 (∞)

θ =  π/2

Arg(z) = π/2.

and

The complex number w = - 2 + 2i

Compare it to x + iy

x = - 2 , y = 2

Since x < 0, use the formula for the argument

θ = tan-1(y/x) + 180o

θ = tan-1(2/-2) + 180o

θ = tan-1(-1) + 180o

θ = - 45 + 180o

θ = - π/4 + π

θ = 3π/4

Arg(w) = 3π/4.

Question 4)

Apply integral on each side.

When x and y = 1

Question 5)

Apply the formula

And

In this case assume the area A is above the x axis and so we can simply evaluate the integral between required limits.

Area is 20.25 square units.

Question 6)

Apply the limits.

Question 7)

The curve y = 3x(1/3)

Differentiating  on each side with respect of x .

y' = 3[1/3 x(1/3)-1]

y' = x(-2/3)

Substitute the values (x , y ) = (1, 3) in above equation.

y' = 1(-2/3)

y' = 1

This is the slope of tangent line to the curve at (1, 3).

To find the tangent line equation, substitute the values of m = 1 and (x, y ) = (1, 3).  in the slope intercept form of an equation y = mx + b.

3 = 1(1) + b

3 - 1= b

b = 2

Substitute m = 1 and b = 2 in y = mx + b

y = x + 2

Tangent line is y = x + 2.