+0

# Complex number

0
258
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Suppose that a and b are positive integers such that (a-bi)^2 = 8 - 6i + 13 - 4i. What is a-bi?

Apr 12, 2022

#1
+30
+4

We can mutiply out (a-bi)2 to get a2 - 2abi - b2.

Since a and b are positive integers, the -10i on the right side of the equation must come from -2abi.

Therefore -10i = -2abi so ab = 5.

And now we see this problem is literally impossible because 5 can only factor into 1 and 5 or 5 and 1, and (1 - 5i)2 is -24 - 10i while (5 - i)2 is 24 - 10i. So unless I'm dumb and missed something, which wouldn't suprise me, this question is either has a typo or is impossible.

Assuming it's a typo and the left side is actually adds to 24 - 10i and not 21 - 10i, the answer is 5 - i.

Apr 12, 2022

#1
+30
+4

We can mutiply out (a-bi)2 to get a2 - 2abi - b2.

Since a and b are positive integers, the -10i on the right side of the equation must come from -2abi.

Therefore -10i = -2abi so ab = 5.

And now we see this problem is literally impossible because 5 can only factor into 1 and 5 or 5 and 1, and (1 - 5i)2 is -24 - 10i while (5 - i)2 is 24 - 10i. So unless I'm dumb and missed something, which wouldn't suprise me, this question is either has a typo or is impossible.

Assuming it's a typo and the left side is actually adds to 24 - 10i and not 21 - 10i, the answer is 5 - i.

justchillin Apr 12, 2022
#2
+118574
+1

(a-bi)^2 = 8 - 6i + 13 - 4i

$$(a-bi)^2 = 8 - 6i + 13 - 4i\\ a^2-b^2-2abi=21-10i\\ a^2-b^2=21\qquad and \qquad -2abi=-10i\\ a^2-b^2=21\qquad and \qquad ab=5\\$$

As just chillin said, there are not integer solution set for this.

Apr 12, 2022