I have an exam tomorrow and our teacher gave us some revision questions on complex numbers. There were three that I could not do and we did not have time in class to solve them. Here they are:
1) When does | z + w | = | z | + | w |?
2) Solve 2iz^2 - (3+4i)z + 5 = 0.
3)When is z + iz = iz^2 + i^2?
In all of the above questions, z = a + bi and w = c + di.
Easy to understand explanations would be appreciated. Thanks heaps!! =D
1) When does | z + w | = | z | + | w |?
2) Solve 2iz^2 - (3+4i)z + 5 = 0.
3)When is z + iz = iz^2 + i^2?
In all of the above questions, z = a + bi and w = c + di.
Easy to understand explanations would be appreciated. Thanks heaps!! =D
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1)Plot point z and -w in the Argand diagram at points A and B then BA=z+w.
In triangle OAB ( O is the origin), the length of any side cannot be greater than the sum of the
lengths of the other two sides so |z+w|<=|z|+|-w|=|z|+|w|. The equality can occur only when
w, z and O are in a straight line so w=cz where c is a real constant.
2)Sub z=a+ib, where a and b are real, multiply out. Re and Im parts are both zero and this
will give two equations for a and b which you solve simultaneously.Not very nice!
Another way is to use the quadratic formula which will lead to finding the square root of the complex
number.-7-16i and this is not exact. I suspect there is a typo in the equation.
3) Write as iz^2-(1+i)z-1=0 => Formula gives z=[(1+i)-sqrt((1+i)^2+4i)]/(2i) or [(1+i)+sqrt((1+i)^2+4i)]/(2i)
(1+i)^2+4i=6i and sqrt(6i)=+(sqrt3+isqrt3) or -(sqrt3+isqrt3) and you can then obtain the two values of z.
I think Q2 and Q3 have too much awkward algebra to be good tests of the principles.
Q1 is OK if you are aware of the triangle inequality but is very tricky otherwise.
In triangle OAB ( O is the origin), the length of any side cannot be greater than the sum of the
lengths of the other two sides so |z+w|<=|z|+|-w|=|z|+|w|. The equality can occur only when
w, z and O are in a straight line so w=cz where c is a real constant.
2)Sub z=a+ib, where a and b are real, multiply out. Re and Im parts are both zero and this
will give two equations for a and b which you solve simultaneously.Not very nice!
Another way is to use the quadratic formula which will lead to finding the square root of the complex
number.-7-16i and this is not exact. I suspect there is a typo in the equation.
3) Write as iz^2-(1+i)z-1=0 => Formula gives z=[(1+i)-sqrt((1+i)^2+4i)]/(2i) or [(1+i)+sqrt((1+i)^2+4i)]/(2i)
(1+i)^2+4i=6i and sqrt(6i)=+(sqrt3+isqrt3) or -(sqrt3+isqrt3) and you can then obtain the two values of z.
I think Q2 and Q3 have too much awkward algebra to be good tests of the principles.
Q1 is OK if you are aware of the triangle inequality but is very tricky otherwise.
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1) | z + w | = | z | + | w |
sqrt((a + c)^2 + (b + d)^2) = sqrt(a^2 + b^2) + sqrt(c^2 + d^2)
(a + c)^2 + (b + d)^2 = a^2 + b^2 + c^2 + d^2 + 2sqrt((a^2 + b^2)(c^2 + d^2))
a^2 + c^2 + 2ac + b^2 + d^2 + 2bd = a^2 + b^2 + c^2 + d^2 + 2sqrt((a^2 + b^2)(c^2 + d^2))
ac + bd = sqrt((a^2 + b^2)(c^2 + d^2))
a^2c^2 + b^2d^2 + 2abcd = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2
2abcd = a^2d^2 + b^2c^2
sqrt((a + c)^2 + (b + d)^2) = sqrt(a^2 + b^2) + sqrt(c^2 + d^2)
(a + c)^2 + (b + d)^2 = a^2 + b^2 + c^2 + d^2 + 2sqrt((a^2 + b^2)(c^2 + d^2))
a^2 + c^2 + 2ac + b^2 + d^2 + 2bd = a^2 + b^2 + c^2 + d^2 + 2sqrt((a^2 + b^2)(c^2 + d^2))
ac + bd = sqrt((a^2 + b^2)(c^2 + d^2))
a^2c^2 + b^2d^2 + 2abcd = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2
2abcd = a^2d^2 + b^2c^2