# Complex NumbersWatch

#1
Find in the form a + ib.

After giving up and looking at the answer, it gave:

How did they derive that?
0
quote
9 years ago
#2
arccos(4) = a + ib so 4 = cos(a + ib); expand using compound angle formula, use the hyperbolic identities for cos(ib) and sin(ib) and equate real and imaginary parts.
quote
#3
(Original post by Glutamic Acid)
arccos(4) = a + ib so 4 = cos(a + ib); expand using compound angle formula, use the hyperbolic identities for cos(ib) and sin(ib) and equate real and imaginary parts.
Works like a charm - thank's GA
0
quote
#4
The second part of the question is tough:

Express in terms of a + ib

So:

comparing real & imaginary:

(*)
(**)

Considering (**)
the only way that sinhb can form 0 is if b=0 which will mess up the imaginary part => cosa = 0 and so only

we must wait from deductions from (*) to fine-tune our values of 'a'.

Considering (*)

a cannot be the -1's which sina produces =>

b is simply [which is the correct answer]

However, my answer for 'a' is wrong; it should be:

0
quote
9 years ago
#5
Your answer for a is correct, just take pi/2 out as a factor and you'll see.
quote
#6
(Original post by ghostwalker)
Your answer for a is correct, just take pi/2 out as a factor and you'll see.
Darn it

Thanks ghostwalker
0
quote
X

new posts

Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of Lincoln
Mini Open Day at the Brayford Campus Undergraduate
Wed, 19 Dec '18
• University of East Anglia
Fri, 4 Jan '19
• Bournemouth University
Wed, 9 Jan '19

### Poll

Join the discussion

Yes (68)
77.27%
No (20)
22.73%