Four complex numbers form the vertices of a square in the complex plane. Three of the numbers are $-19 32i,$ $-5 12i,$ and $-22 15i$. What is the fourth number

If the three numbers of a square in the complex plane are [tex]-19+32i,-5+12i and -22+15i[/tex] , then the fourth complex number [tex]-2+19i[/tex].

Given [tex]-19+32i,-5+12i and -22+15i[/tex] are three numbers.

Complex numbers are those numbers which extends the real numbers with an imaginary i. In this [tex]i^{2}=-1[/tex]. Major complex numbers are in the form a+ bi where a and b are real numbers.

let the fourth complex numbers be [tex]x+yi[/tex]. Then according to question;

=[tex](-22+15i)-(-5+12i)[/tex]

=(cos π/2+i sin π/2) [tex](x+yi)-(-5+12i)[/tex]

[tex]-17+3i=-y+12[/tex][tex]+(x+5)i[/tex]

Now solving for x and y by equating both sides.

x=-2 and y=29

Put the value of x and y in [tex]x+yi[/tex]

Z=-2+29i

Hence if the three numbers which forms vertices of a square are [tex]-19+32i,-5+12i,-22+25i[/tex] then the fourth complex numbers be [tex]-2+29i[/tex].

Learn more about complex numbers at https://brainly.com/question/10662770

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