Question

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

Answer 1

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

Given $-19+32i,-5+12i and -22+15i$ are three numbers.

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

let the fourth complex numbers be $x+yi$. Then according to question;

=$(-22+15i)-(-5+12i)$

=(cos π/2+i sin π/2) $(x+yi)-(-5+12i)$

$-17+3i=-y+12$$+(x+5)i$

Now solving for x and y by equating both sides.

x=-2 and y=29

Put the value of x and y in $x+yi$

Z=-2+29i

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

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

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