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Q. If for a complex number, `z=x+iy ,` `sec^-1(z-2)/i` is an acute angle ,then A)...

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user8235304 | Student, Grade 11 | Valedictorian

Posted August 18, 2013 at 11:39 AM via web

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Q. If for a complex number, `z=x+iy ,` `sec^-1(z-2)/i` is an acute angle ,then

A) x=2,y=1

B) x<2,y<-1

C) xy<0

D) x=2,y>1

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 18, 2013 at 12:27 PM (Answer #1)

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You need to evaluate first the argument of the function inverse `sec^(-1) (z - 2)/i` , such that:

`(z - 2)/i = -i*(z -2)/(-i*i) => (z - 2)/i = -i*(z -2)/(-i^2)`

Since `i^2 = -1` yields:

`(z - 2)/i = -i*(z -2)/1 = -i*(z -2)`

Performing the multiplication yields:

`(z - 2)/i = -z*i + 2i`

Replacing `x + i*y` for `z` yields:

`(z - 2)/i = -(x + i*y)*i + 2i`

`(z - 2)/i = -x*i - i^2*y + 2i`

`(z - 2)/i = -x*i + y + 2i => (z - 2)/i = y + i*(2 - x)`

Since the function `sec theta = 1/cos theta` yields that theta is an acute angle if `cos theta > 0` , hence, `0 < theta < pi/2` .

Reasoning by analogy, the angle `theta = sec^(-1)(y + i*(2 - x))` is acute if the argument `y + i*(2 - x)` make the following inequality to hold, such that:

`y + i*(2 - x) > 1 =>y + i*(2 - x) > 1=> {(y>1),(2 - x = 0)} => {(y > 1),(x = 2)}`

Hence, checking what is the valid answer, under the given conditions, yields `D) x = 2, y>1.`

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