In the 2nd method, you need to be careful when applying the rule √(z/w)=√z/√w, because it doesn't always hold in the world of complex numbers, assuming (as you do) that √z represents the principal square root. For example, taking z=1 and w=-1, we get √(z/w)=√(1/-1)=√(-1)=i but √z/√w=√1/√(-1)=1/i=-i, so in fact √(1/-1)=-√1/√(-1) To see what's going on here, it helps to use polar form. We have z=1=cis 0, w=-1=cis π (using the principal arguments). So z/w=1/-1=cis 0/cis π=cis(-π). Taking principal square roots, √z=cis(0/2)=cis 0, √w=cis(π/2), so √z/√w=cis(0-π/2)=cis(-π/2) On the other hand, √(z/w)=√cis(-π)=cis(-π/2) and all seems to be fine. But wait! cis(-π) is not written using its principal argument (-π is just outside the allowed range -π
My method was setting the expression equal to a+bi and squaring both sides to get rid of the radical. Then I took the complex conjugate, and from there I got the two square roots of 2i, which are 1+i and -1-i.
That square root sign is single valued, which is why he said "principle value". Frankly, that sign should be banned as it causes lots of confusion when dealing with higher power roots.
@kevinmadden1645 You can only have one! If it was a real number, say 4, both 2 and -2 squared are solutions. But by convention, 2 is the principal root. Which is why we put "+/-" before the square root sign if we mean both. I don't know who decided that the solution with the positive real value is the principal root but Syber said that was the convention he was following.
When I continue: Take out sqrt(2) and the fraction becomes (-4+3i)/(3+4i) Now multiply top and btm by -i to get Sqrt (-2/i) sqrt,(3+4i)/sqrt(3+4i) = sqrt(2i) = (1+i) Similar to method 2.
In the 2nd method, you need to be careful when applying the rule √(z/w)=√z/√w, because it doesn't always hold in the world of complex numbers, assuming (as you do) that √z represents the principal square root.
For example, taking z=1 and w=-1, we get √(z/w)=√(1/-1)=√(-1)=i
but √z/√w=√1/√(-1)=1/i=-i,
so in fact √(1/-1)=-√1/√(-1)
To see what's going on here, it helps to use polar form.
We have z=1=cis 0, w=-1=cis π (using the principal arguments).
So z/w=1/-1=cis 0/cis π=cis(-π).
Taking principal square roots,
√z=cis(0/2)=cis 0, √w=cis(π/2),
so √z/√w=cis(0-π/2)=cis(-π/2)
On the other hand,
√(z/w)=√cis(-π)=cis(-π/2)
and all seems to be fine.
But wait! cis(-π) is not written using its principal argument (-π is just outside the allowed range -π
My method was setting the expression equal to a+bi and squaring both sides to get rid of the radical. Then I took the complex conjugate, and from there I got the two square roots of 2i, which are 1+i and -1-i.
You are good! 😍
Sir, little bit actually,
sqrt(3 + 4i)
= sqrt(4 - 1 + 4i)
= sqrt(2² + i² + 2×2×i)
= (2 + i)
and not (1 + 2i) 🥹
sqrt(-8 + 6i) / sqrt(3 + 4i)
= (1 + 3i) / (2 + i)
= (1 + 3i)(2 - i) / (2 + i)(2 - i)
= (2 - i + 6i + 3) / 5
= (5 + 5i) / 5
= (1 + i)😊
Continue! Sqrt(2i) equals 1+I and -1-i.
That square root sign is single valued, which is why he said "principle value".
Frankly, that sign should be banned as it causes lots of confusion when dealing with higher power roots.
@mcwulf25 Why is 1+I the principal( not principle) square root of 2i? Both 1+I and -1-i equal 2i when squared.
@kevinmadden1645 You can only have one! If it was a real number, say 4, both 2 and -2 squared are solutions. But by convention, 2 is the principal root. Which is why we put "+/-" before the square root sign if we mean both. I don't know who decided that the solution with the positive real value is the principal root but Syber said that was the convention he was following.
I got 2i.
Your method seems too complicated.
Write expression as x = sqrt(2(3i-4)/(3+4i))
Notice that 3 + 4i = 1/i * (3i - 4).
So x = sqrt(2/(1/i) = sqrt(2i) = sqrt(2) * sqrt(i)
Simplify sqrt(i) using i = e^i(pi/2 + 2n*pi) => i^1/2 = e^i(pi/4 + n*pi) = e^i*pi/4 * e^i*n*pi = sqrt(2)/2 * (1+ i) * (-1)^n
So x = (1+ i) * (-1)^n = 1+i or -1-i.
Not seen the video yet but I notice the modulus of the numerator is 10 and the denominator is 5.
When I continue: Take out sqrt(2) and the fraction becomes (-4+3i)/(3+4i)
Now multiply top and btm by -i to get
Sqrt (-2/i) sqrt,(3+4i)/sqrt(3+4i)
= sqrt(2i)
= (1+i)
Similar to method 2.
2i or not 2i😂😂😂
I got u! 😁
rewrite as √( (numer) / (denom) )
numer = -8+6i
denom = 3+4i
note that numer = 2i * denom (or do the division)
almost donre. The answer = √(2i)
= (- 8 + 6i)/(3 + 4i)
= 2.(- 4 + 3i).(3 - 4i)/[(3 + 4i).(3 - 4i)]
= 2.(- 12 + 16i + 9i - 12i²)/[9 - 16i²] → where: i² = - 1
= 2.(- 12 + 25i + 12)/[9 + 16]
= 2.(25i)/[25]
= 2i
= [√(- 8 + 6i)] / [√(3 + 4i)]
= √[(- 8 + 6i) / (3 + 4i)]
= √(2i)