Complex Analysis: Find 2 Square Roots, Solve Eqn, Form Triangle

[mtayab1994]

Homework Statement

1- Find the two square roots of the complex number z=3+4i.

2a- Solve in ℂ the equations: (E): 4z^2-10iz-7-i=0

b- Let a and b be solutions to (E) such that: Re(a)<0 and the two points A and B plots/pictures of a and b. Show that b/a=1-i. Conclude that AOB is an equilateral triangle.

The Attempt at a Solution

1- After solving (p+qi)^2=3+4i i found that the solutions were either 2+i or -2-i.

2-a For the complex equation i found two complex roots: z1=(-3+6i)/8 and z2=(3+14i)/8.

b- So i took the two solutions that i found from the previous question and chose a=z1 and b=z2 and after computing i got a whole different answer. Is my work correct, if not some help would be very much appreciated.

 

[haruspex]

2-a For the complex equation i found two complex roots: z1=(-3+6i)/8 and z2=(3+14i)/8.

I get a different result. Pls post your working.

 

[HallsofIvy]

Homework Statement

1- Find the two square roots of the complex number z=3+4i.

2a- Solve in ℂ the equations: (E): 4z^2-10iz-7-i=0

b- Let a and b be solutions to (E) such that: Re(a)<0 and the two points A and B plots/pictures of a and b. Show that b/a=1-i. Conclude that AOB is an equilateral triangle.

The Attempt at a Solution

1- After solving (p+qi)^2=3+4i i found that the solutions were either 2+i or -2-i.

2-a For the complex equation i found two complex roots: z1=(-3+6i)/8 and z2=(3+14i)/8.

If z= (-3+ 6i)/8 then z^2= [(9- 36)- 2(18i)]/64= -27/64- (9/8)i so 4z^2- 10iz- 7- i= -27/4- (9/2)i+ (15/4)i+ 15/2- 7- i= (-27/4+ 15/2- 7)+ (15/4- 9/2- 1)i= (-27+ 30- 28)/4+ (15/4- 18/4- 4/4)i= -25/4- (7/4)i, NOT 0.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> b- So i took the two solutions that i found from the previous question and chose a=z1 and b=z2 and after computing i got a whole different answer. Is my work correct, if not some help would be very much appreciated. </div> </div> </blockquote>

 

[mtayab1994]

Sorry I was wrong on the roots of the equation they are correct now i got:

z1=(1/2)+(3i/2) and z2=(-1/2)+i and that certainly gives 1-i when you take z2=a and z1=b.

But the conclusion i can't quite fathom, any help with that please.

 

[haruspex]

But the conclusion i can't quite fathom, any help with that please.

What will the third side look like as a complex number (in terms of a and b)? What will its ratios be to the other two?