Question

If z=a+bi for real numbers a and b and z^2=21-201, what is the value of |a|+|b|?

Answer 1

z=a+jb

Answer 2

you can use de moivre's theorem

Answer 3

agg, sorry
it's actually z^2=21-20i

Answer 4

\[z=(-b \pm \sqrt{b^2 - 4ac})/2a\]

Answer 5

yh, re-arrange to give z^2+i20-21

Answer 6

actually note that (a+bi)^2 = (a^2 - b^2) + 2abi.
Now set that equal to 21-20i and solve for a and b.

Answer 7

use the almighty formula , identify a, b and c and subs.t into the formula

Answer 8

@ JamesJ you do realize that that is a confusing method as per not related to complex numbers

Answer 9

ah...thanks!

Answer 10

What do you mean? It's absolutely germane to complex numbers.

Answer 11

In fact, try and use your method and you'll get stuck.

Answer 12

interesting, i might have not come across it

Answer 13

\[z=a+bi \implies z^2=(a+bi)^2=a^2+2abi-b^2=21-20i \implies (a^2-b^2)=21\]
\[2ab=-20\]
Now solve. Two equations two unknowns.

Answer 14

That first line should read:
\[a^2-b^2=21^*\]

Answer 15

wait, JamesJ, the a and b formula you put up there doesn't work...or am i wrong?

Answer 16

As James said, don't use QF. lolz

Answer 17

oh...never mind! :P

Answer 18

mal19 has laid it out for you. Two complex numbers are equal if and only if their real and imaginary components are equal.

Hence if z = a+bi and z^2 = 21 - 20i, it must be that:
\[ a^2 - b^2 = 21 \ \ \ \ 2ab = -21 \]

Answer 19

Yup :)

Answer 20

**second equation should be = -20

Answer 21

but how do I get to the absolute value?

Answer 22

Solve for a and b. Then take the absolute value at the very end.

Answer 23

... I got something that's definitely not an answer(which is going to be between 3 and 7, inclusive)

Answer 24

Using the second equation, 2ab = -20, then b = -10/a. Substitute that into the equation
a^2 - b^2 = 21 and solve for a.

Take those solutions and find the corresponding b.

Answer 25

So we have:
\[2ab=-20 \implies b=\frac{-10}{a}\]
But:
\[a^2-b^2=21\] as well so:
\[a^2- \left( \frac{-10}{a} \right)^2=21 \implies a^2-\frac{100}{a^2}=21\]
Multiply by a^2 to give:
\[a^4-100=21a^2 \implies a^4-21a^2-100=0\]
Make the substitution y=a^2 and we get:
\[y^2-21y-100=0 \implies (y-25)(y+4)=0 \implies (a^2-25)(a^2+4)=0 \implies a= \pm 5\]
\[ a= \pm 2i\]
Unless I'm doing something completely wrong, lol. Now solve b?

Answer 26

after the implies it should be:
\[a= \pm 5\]

Answer 27

Right. a must be real, so the values of a are +5 or -5. Find now the corresponding values of b.

Answer 28

And I'm assuming you only care about the real answer (or else the number wouldn't be in the form a+bi (think, if a=2i then you have (2+b)i, which is different) So a=5 (its an abs value so you don't need to worry about sign)

Answer 29

Exactly^

Answer 30

the answer is 7, right?
thanks everyone!!! (If I could medal multiple times, I would)

can you guys stick around? I have other questions.
I swear I'm seriously stuck on these (I'm not lazy)
I'd really like to know how to solve them,(our teacher gave them as "extra practice" over the break, but he's not talking about then in class)

Answer 31

Correct, if my mental arithmetic is correct.

Answer 32

By the way, the two solutions 5 - 2i and -5 + 2i are additive opposites, which is exactly what we would expect.

You can and should verify that they do indeed square to 21 - 20i.

Answer 33

Post the new question on the left, thanks.

Answer 34

i guess we learn something new everyday. I'm quite sure my method would give the same answer

Answer 35

i can show you and i won't get stuck ^_^

Answer 36

\[z^2=21-i20=>z=\sqrt{21-i20}\]

Answer 37

\[z=(21-i20)^{1/2}\]

Answer 38

You aren't solving z...You're solving a and b... Now explain how to find the sqrt of that number as well please :)

Answer 39

yh, can't we find the two solutions of z and then find the magnitude of a and b from those two solutions ? for ex. if z1=a1+ib1 and z2=a2+ib2

Answer 40

right, exactly. Where do you go from there.

Answer 41

we will need to transform that into polar form

Answer 42

\[r<\theta\]

Answer 43

Given an arbitrary complex number z and w multiplying them would give:
\[z*w=rq(\cos(\theta+\phi)+isin(\theta+\phi))\]
Assuming a modulus of r and q for z and w respectively. How would that help?

Answer 44

Also, my steps (way) above make the assumption that:
\[a \ne 0\] (I believe we have to make that assumption***)

Answer 45

yh, i meant..

Answer 46

\[z ^{-n}=r ^{-n}(\cos(n \theta)-isin(n \theta))\]

Answer 47

\[z=(21-i20)^{1/2}=(29<44)^{1/2}\]

Answer 48

\[z=(29^{1/2})<44/2)\]

Answer 49

What are you doing? lol

Answer 50

i'm trying to find the two roots to the equation

Answer 51

lol

Answer 52

Okay, go for it :P

Answer 53

you said you me to solve z=(21-j20)^1/2

Answer 54

you said you me to solve z=(21-j20)^1/2\[z=(21-j20)^{1/2}\]

Answer 55

i do know \[z _{1}=5.4<-22^{o}, z _{2}=5.4<158^{o}\]

Answer 56

Degrees!

You should carry the arctan(-20/21) and then work out the cos and sin of half that quantity (+ pi) so you recover the exact solutions.

Answer 57

\[\therefore z _{1}=5.4(\cos(-22)+jsin(-22)) = 4.6-j2\]

Answer 58

yh, you can't add pi if x>0

Answer 59

\[z _{2} = -5+j2\]

Answer 60

as you can see we both arrived at the same ans