Answer to Question #117769 in Algebra for Samuel

Question

Given that z=1+i√z ,express in the form a+ib each of the complex numbers p=z+1/z ,q=z-1/z . In an Argand diagram ,P and Q are the points which represent p and q respectively . O is the origin , M is the midpoint of PQ and G is the point on OM such that OG=two-thirds of OM . Prove that angle PGQ is a right angle . Please I need the answers as soon as possible . Thanks

Answer 1

Given, z = 1 + i$\sqrt{\smash[b]{2}}$

p = z + 1/z

p = 1 + i$\sqrt{\smash[b]{2}}$ + 1/(1 + i$\sqrt{\smash[b]{2}}$ )

p = 1 + i$\sqrt{\smash[b]{2}}$ + (1 - i$\sqrt{\smash[b]{2}}$ )/((1 + i$\sqrt{\smash[b]{2}}$ ) × (1 - i$\sqrt{\smash[b]{2}}$ ))

p = 1 + i$\sqrt{\smash[b]{2}}$ + (1 - i$\sqrt{\smash[b]{2}}$ )/(1² - (i$\sqrt{\smash[b]{2}}$ )²)

p = 1 + i$\sqrt{\smash[b]{2}}$ + (1 - i$\sqrt{\smash[b]{2}}$ ​ )/(1 + 2)

p = 1 + i$\sqrt{\smash[b]{2}}$ ​ + (1 - i$\sqrt{\smash[b]{2}}$ ​ )/3

p = (1 + 1/3) + i($\sqrt{\smash[b]{2}}$ - ($\sqrt{\smash[b]{2}}$ ​/3))

Therefore, p = (4/3) + i(2$\sqrt{\smash[b]{2}}$/3), where a = (4/3) and b = (2$\sqrt{\smash[b]{2}}$​/3).

q = z - 1/z

q = 1 + i$\sqrt{\smash[b]{2}}$ ​ - (1/(1 + i$\sqrt{\smash[b]{2}}$ ​ ))

q = 1 + i$\sqrt{\smash[b]{2}}$ ​ - ((1 - i$\sqrt{\smash[b]{2}}$ ​ )/(1 + i$\sqrt{\smash[b]{2}}$ )(1 - i$\sqrt{\smash[b]{2}}$ ​ ))

q = 1 + i$\sqrt{\smash[b]{2}}$ ​ - (1 - i$\sqrt{\smash[b]{2}}$ ​ )/3

q = (1 - (1/3)) + i($\sqrt{\smash[b]{2}}$ ​ + ($\sqrt{\smash[b]{2}}$ ​/3))

Therefore, q = (2/3) + i(4$\sqrt{\smash[b]{2}}$ ​/3), where a = (2/3) and b = (4$\sqrt{\smash[b]{2}}$ ​/3).

Since P and Q represents the points of p and q,

∴ P = (4/3 , 2$\sqrt{\smash[b]{2}}$ /3) & Q = (2/3 , 4$\sqrt{\smash[b]{2}}$ /3)

Given that M is the midpoint of PQ.

Let M = (x,y).

Therefore,

x = ((4/3) + (2/3))/2 & y = ((2$\sqrt{\smash[b]{2}}$ /3) + (4$\sqrt{\smash[b]{2}}$ /3))/2

x = 3/3 = 1 & y = 3$\sqrt{\smash[b]{2}}$ /3 = $\sqrt{\smash[b]{2}}$

M = (3/3 , 3$\sqrt{\smash[b]{2}}$ /3)

Distance OM = $\sqrt{\smash[b]{(1 - 0)² + (\sqrt{\smash[b]{2}} - 0)²}}$

OM = $\sqrt{\smash[b]{1 + 2}}$

OM = $\sqrt{\smash[b]{3}}$

Given that, OG = (2/3) × OM

Therefore, OG = 2$\sqrt{\smash[b]{3}}$ /3

Let the slope of line OM be l, which is calculated as below :

l = $(\sqrt{\smash[b]{2}} - 0)$ /(1 - 0)

l = $\sqrt{\smash[b]{2}}$

Since, GM lies on OM, hence the slope of GM is the same as the slope of OM, i.e., l.

Let coordinates of G = (s,t)

Slope of GM = l

(t - $\sqrt{\smash[b]{2}}$ )/(s - 1) = $\sqrt{\smash[b]{2}}$ /1

(t - $\sqrt{\smash[b]{2}}$ ) = $\sqrt{\smash[b]{2}}$ & (s - 1) = 1

t = 2$\sqrt{\smash[b]{2}}$  & s = 2

Since GM is smaller than OM and it lies on OM and not outside it, hence we multiply the two coordinates by 1/3 to bring it to scale.

Therefore, t = 2$\sqrt{\smash[b]{2}}$ /3 & s = 2/3.

i.e., G = (2/3 , 2$\sqrt{\smash[b]{2}}$ /3 )

Distance PG =  $\sqrt{\smash[b]{((4/3) - (2/3))² + (2\sqrt{\smash[b]{2}}/3 - 2\sqrt{\smash[b]{2}}/3)² }}$

PG = $\sqrt{\smash[b]{(2/3)²}}$

PG = 2/3

Distance QG = $\sqrt{\smash[b]{(2/3 - 2/3)² + ((4\sqrt{\smash[b]{2}}/3) - (2\sqrt{\smash[b]{2}}/3))²}}$

QG = $\sqrt{\smash[b]{(2\sqrt{\smash[b]{2}}/3)²}}$

QG = $2\sqrt{\smash[b]{2}}/3$

Distance PQ = $\sqrt{\smash[b]{((2\sqrt{\smash[b]{2}}/3) - (4\sqrt{\smash[b]{2}}/3))² + ((4/3) - (2/3))² }}$

PQ = $\sqrt{\smash[b]{(8/9) + (4/9)}}$

PQ = $\sqrt{\smash[b]{(12/9)}}$

PQ = 2$\sqrt{\smash[b]{3}}/3$

PG² + QG² = (2/3)² + (2$\sqrt{\smash[b]{2}}$ /3 )²

PG² + QG² = (4/9) + (8/9) = 12/9

PG² + QG² = (2$\sqrt{\smash[b]{3}}$ /3)²

PG² + QG² = PQ²

The above equation is the equation of pythagoras theorem for a right angled triangle, i.e., a² + b² = c², where c is the hypotenuse and a & b are the sides making the right angle.

Therefore, PGQ is a right angled triangle with PQ as the hypotenuse. The angle opposite to the hypotenuse is the right angle.

Therefore, angle PGQ is a right angle.