Complex Numbers Question - Patel (1 Viewer)

[ashokkumar]

Can someone show me how to do these?

If z1 = 1 + sqrt3*i and z2 = sqrt3 + i, find:

a) z1 + z2 in mod-arg form
b) z1 - z2 in mod-arg form
c) (z1 + z2)/(z1 - z2) in the form A + iB
d) [(z1 + z2)/(z1 - z2)]^2 in the form A + iB

Answers:
a) sqrt2*(1+sqrt3)cis(pi/4)
b) sqrt2*(sqrt3-1)cis(3pi/4)
c) -(2+sqrt3)i
d) -7 -4sqrt3

 

[biopia]

z1 + z2 = (1 + √3) + (√3 + 1)i
Modulus = √[(1 + √3)² + (1 + √3)²]
= √2(1 + √3)²
= √2*(1 + √3)
Argument: tan@ = (1 + √3)/(√3 + 1) = 1
@ = π/4
Therefore z1 + z2 = √2*(1 + √3)cis(π/4)

z1 - z2 = (1 - √3) + (√3 - 1)i
Modulus = Same as before except (-) instead of (+)
Argument: Someone can probably explain this better, but I think about it geometrically. I think about where the argument has to be based on the quadrant that z1 and z2 are in individually, and tan@ will still equal 1, so @ has to be = 3π/4

The last two I can't be bothered doing lol, but one way of approaching them is to write it out and rationalising the denominator. They'll be messy because you will have a lot of 'the difference of two squares' going on, but other than that, be relatively easy. Sorry I can't be of more help.

 

[biopia]

Hey again...

I realised there was a WAY better way to do the last two parts lol. Instead of trying to rationalise the denominator, just divide in mod arg form and then convert into A + iB

So:
[√2*(1 + √3)cis(π/4)]/[√2*(1 - √3)cis(3π/4)]
In this case, we divide the modulus and subtract the arguments.
[(1 + √3)/(1 - √3)]cis(-π/2)

[(1 + √3)/(1 - √3)] = -(2 + √3) {Rationalise THIS denominator lol}

cos(-π/2)=0
sin(-π/2)=-1

= -(2 + √3)cos(-π/2) + -(2 + √3)isin(-π/2)
= -(2 + √3)i

d is much the same just use demoivre's theorem to square each complex number in mod arg form and then approach the division in the same way.