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July 31, 2014

July 31, 2014

Posted by **Courtney** on Wednesday, February 20, 2008 at 5:50pm.

15-4i and a-ai where a is greater than 0

- calculus -
**Damon**, Wednesday, February 20, 2008 at 6:10pmif z = a + b i

r^2 = sqrt( a^2 + b^2)

and a = r cos T

and b = r sin T

so

if a = 15

and b = -4

then r^2 = 15^2 + (-4)^2 = 225+16

r = 15.524

cos T = 15/15.524

so T = 14.9 or -14.9

sin T = -4/15.52

so T = -14.9 or (180-14.9)=165.1

T = -14.9 which is +345.1 in the fourth quadrant satisfies both, so that is it. (Besides we could see that 15-4i was in quadrant 4)

r = sqrt (2 a^2) = a sqrt 2

fourth quadrant again

so

sin T = -1/sqrt (2) = = -.707

T = -45 degrees which is + 315 degrees

- calculus -
**Courtney**, Wednesday, February 20, 2008 at 6:17pmim sorry they were two different questions you have to find the modulus and the argument

ex.15-4i

modulus-4

but i dont know what the argument is

- calculus -
- calculus -
**Damon**, Wednesday, February 20, 2008 at 6:34pmI did them as two different questions

for #1

modulus = r = 15.524

argument = T (for theta) = -14.9 degrees (or +345.1 degrees)

for #2

modulus = a sqrt 2

theta = argument = -45 (or 315) degrees

- calculus -
**Courtney**, Wednesday, February 20, 2008 at 6:37pmthanks

- calculus -

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