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I would be very grateful is someone could show me how to complete the following questions:

1) Find the modulus and the argument, in radians in terms of pi, of:

z1 = (1 + i)/(1 -i)

z2 = √2/(1 -i)

Plot z1, z2 and z1 + z2 on a Argand diagram.

I can do these parts. However the question then goes on:

Prove that tan(3pi/8) = 1 + √2

I have calculated that the principal argument of z1 + z2 is arctan(1 + √2) so all I need to show is that the angle is equal to 3pi/8 but I do not know how to do this.

2) Given that z = cos x + isin x show that:

2/(1 + z) = 1 - itan (x/2)

1) Find the modulus and the argument, in radians in terms of pi, of:

z1 = (1 + i)/(1 -i)

z2 = √2/(1 -i)

Plot z1, z2 and z1 + z2 on a Argand diagram.

I can do these parts. However the question then goes on:

Prove that tan(3pi/8) = 1 + √2

I have calculated that the principal argument of z1 + z2 is arctan(1 + √2) so all I need to show is that the angle is equal to 3pi/8 but I do not know how to do this.

2) Given that z = cos x + isin x show that:

2/(1 + z) = 1 - itan (x/2)

the_anomaly

I would be very grateful is someone could show me how to complete the following questions:

1) Find the modulus and the argument, in radians in terms of pi, of:

z1 = (1 + i)/(1 -i)

z2 = √2/(1 -i)

Plot z1, z2 and z1 + z2 on a Argand diagram.

I can do these parts. However the question then goes on:

Prove that tan(3pi/8) = 1 + √2

I have calculated that the principal argument of z1 + z2 is arctan(1 + √2) so all I need to show is that the angle is equal to 3pi/8 but I do not know how to do this.

2) Given that z = cos x + isin x show that:

2/(1 + z) = 1 - itan (x/2)

1) Find the modulus and the argument, in radians in terms of pi, of:

z1 = (1 + i)/(1 -i)

z2 = √2/(1 -i)

Plot z1, z2 and z1 + z2 on a Argand diagram.

I can do these parts. However the question then goes on:

Prove that tan(3pi/8) = 1 + √2

I have calculated that the principal argument of z1 + z2 is arctan(1 + √2) so all I need to show is that the angle is equal to 3pi/8 but I do not know how to do this.

2) Given that z = cos x + isin x show that:

2/(1 + z) = 1 - itan (x/2)

For (1) you've spotted the hard part - the remainder is geometric. Note that z1+z2 is at distance 1 from both z1 and z2. This is because

|(z1+z2)-z1|=|z1|=1 and

|(z1+z2)-z2|=|z2|=1.

This means it lies on the bisector of the line connecting z1 and z2. As z1 had argument pi/2 and z2 had pi/4 then z1+z2 has argument 3pi/8.

For (2) just put it into the form a+ib by multiplying top and bottom by the conjugate of the bottom and then write sinx and cosx in terms of half the angle by using the double angle formulas. With some working missing:-

2/(1+z)

= 2{(1+cosx) - i sinx}/{(1+cosx)^2+sin^2 x}

= 4{c^2- i s c}/{4c^2}

= 1 - it

where s = sin(x/2), c= cos(x/2), t = tan(x/2).

the_anomaly

z1 = (1 + i)/(1 -i)

z2 = ?2/(1 -i)

Plot z1, z2 and z1 + z2 on a Argand diagram.

I can do these parts. However the question then goes on:

Prove that tan(3pi/8) = 1 + ?2

I have calculated that the principal argument of z1 + z2 is arctan(1 + ?2) so all I need to show is that the angle is equal to 3pi/8 but I do not know how to do this.

z1 + z2

= (1 + sqrt(2) + i)/(1 - i) . . . (A)

= (1 + sqrt(2) + i)(1 + i)/2

= [sqrt(2) + i(2 + sqrt(2))]/2 . . . (B)

Taking arguments of (A) and (B), arctan[1/(1 + sqrt(2))] + pi/4 = arctan(1 + sqrt(2)).

For any number x > 0, draw a right-angled triangle with non-hypotenuse sides sqrt(x) and 1/sqrt(x). The triangle proves that arctan(1/x) + arctan(x) = pi/2.

So 3pi/4 = 2arctan(1 + sqrt(2)) and arctan(1 + sqrt(2)) = 3pi/8.

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