Comlex numbers

[jpa]

Hi,

I have a problem to even start with this math problem.

The question is:

The dot 0,2 + i and -1 + 4i are three of the corners in a parallelogram in the complex plane. Define the fourth corner?

Is the third dot, in the coordinate origin?

[complex plane]

The dot 0,2 + i and -1 + 4i are three of the corners in a parallelogram.
Here you have defined only two dots and you say that 3 corners are given.
Which one is third? Something is missing in the problem.

[Guest]

Ok, so it is not only me who thinks that it is something missing here.

Anyway could we from the information given come up with the answer. And could it be that the third corner i (0;0) where the two vectors start?

The answer is: -3+31, 3-3i or 1+5i

[Guest]

I meant: -3+3i, 3-3i or 1+5i

[Guest]

Find the algebraic form of the number:

[tex](\frac{\sqrt{3} + i }{ 1 - i\sqrt{3} }) ^{31}[/tex]

[leesajohnson]

-3+3i, 3-3i or 1+5i

[Guest]

It looks to me like the first point (dot) is 0. The second corner is 2+ i and the third is -1+ 4i.

[HallsofIvy]

Find the algebraic form of the number:
[tex]\left(\frac{\sqrt{3}+ i}{1- i\sqrt{3}}\right)^{31}[/tex]

Pretty straight forward. "Rationalize" the denominator by multiplying the numerator and denominator by the complex conjugate of the denominator.

[tex]\frac{\sqrt{3}+ i}{1- i\sqrt{3}}\frac{1+ i\sqrt{3}}{1+ i\sqrt{3}}= \frac{\sqrt{3}+ i+ 3i- \sqrt{3}}{1+ 3}= \frac{4i}{4}= i[/tex].

Now raise that to the 31 power. 31= 7(4)+ 3 and [tex]i^4= 1[/tex] so [tex]i^{31}= (i^4)^7(i^3)= -i[/tex].