# Trig help with complex number?

Write the complex number in trigonometric form. Round the angle to the nearest hundredth of a degree: -2+11i

Note: So far, I have 5√5 cis(?) correct. Can not figure out the degree.

### 8 Answers

- Anonymous1 decade agoBest Answer
You are converting -2 + 11i to polar form.

|-2 + 11i| = √[ (-2)^2 + 11^2 ] = √(4 + 121) = √125 = 5√5

You are correct with that.

Now you need to find the angle between the positive x-axis and (-2, 11). You can calculate the angle between the x-axis and (a, b) by using tan^-1(b / a). In this case, we have:

tan^-1(11 / -2) ≈ -79.695° ≈ 280.305°

But be careful as the negative signs mess around with the calculation (this is fixed by a function called atan2). If you picture (-2, 11), it is in Quadrant II. 280° is in Quadrant IV. In order to get the angle in Quadrant II, you need to subtract 180°. So we have 100.305°. Therefore, -2 + 11i in polar form is:

-2 + 11 = 5√5 cis[ tan^-1(-5.5) ] ≈ 5√5 cis 100.305°

Hope this helps!

- Anonymous1 decade ago
A complex number is given in the form:

z = |z|exp(iθ), or x+iy = Sqrt(x^2+y^2)*exp[i*arctan(y/x)].

So, z = 5Sqrt(5) * exp(iarctan(-11/2)) = 5Sqrt(5)*cis(-arctan(11/2)).

- ComoLv 71 decade ago
tan Ө = 11/2

Ө = 79.7 °

180 ° - 79.7 ° = 100.3 ° (in second quadrant)

| - 2 + 11 i | = (4 + 121 )^(1/2) = 125^1/2 = 5√5

Vector is then :-

5√5 /_ 100.3 ° = 5√5 ( cos 100.3 ° + i sin 100.3 ° )

- 4 years ago
Identities are proved or disproved, but not solved. LHS = sin(2s) = 2sin(s)cos(s) = 2sin(2)sin(π/2 - s) = -2sin(2)sin(s - π/2) = RHS

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- Anonymous1 decade ago
go ask a calculator

- Anonymous1 decade ago
fgd