Question 184328
I'll do the first part to get you started.



First, multiply {{{a+bi}}} and {{{1+i}}} to get:



{{{(a+bi)(1+i)=a+ai+bi+bi^2=a+ai+bi-b}}} (which is what you have)




{{{(a+bi)(1+i)=(a-b)+(a+b)i}}} Rearrange and group like terms



So we essentially have two complex numbers {{{a+bi}}} and {{{(a-b)+(a+b)i}}}



If we plot these numbers on the complex plane, then we'll have the points (a,b) and (a-b,a+b)



Now think of these points as vectors



So we have two vectors *[Tex \LARGE \displaystyle \v{v_{1}}=<a,b>] and *[Tex \LARGE \displaystyle \v{v_{2}}=<a-b,a+b>]



So the goal now is to find the angle between the two vectors (so we'll know the exact angle needed to rotate)



Remember, the angle between two vectors can be found through the formula


*[Tex \LARGE \cos(\theta)=\frac{\v{a}\cdot\v{b}}{\left|\v{a}\right|\left|\v{b}\right|}] where   *[Tex \Large \left|\v{a}\right|] is the magnitude (ie length) of vector *[Tex \Large \v{a}] and *[Tex \Large \left|\v{b}\right|] is the magnitude (ie length) of vector *[Tex \Large \v{b}]



Note: I'm making assumptions that you are familiar with vector calculations (such as the dot product and finding the magnitude of a vector). If my assumptions are wrong, please let me know.



So let's calculate the dot product:




*[Tex \Large \v{v}_{1}\cdot\v{v}_{2}=(a)(a-b)+(b)(a+b)=a^2-ab+ab+b^2=a^2+b^2]




Now let's calculate the magnitudes:


*[Tex \Large \left|\v{v}_{1}\right|=\left|<a,b>\right|=\sqrt{a^2+b^2}] 



*[Tex \Large \left|\v{v}_{2}\right|=\left|<a-b,a+b>\right|=\sqrt{(a-b)^2+(a+b)^2}=\sqrt{a^2-2ab+b^2+a^2+2ab+b^2}=\sqrt{2a^2+2b^2}=\sqrt{2(a^2+b^2)}]


Now multiply the magnitudes


*[Tex \Large \left|\v{v}_{1}\right|\cdot\left|\v{v}_{2}\right|=\sqrt{a^2+b^2}\cdot\sqrt{2(a^2+b^2)}=\sqrt{2(a^2+b^2)(a^2+b^2)}=\sqrt{2(a^2+b^2)^2}=\sqrt{2}\cdot\sqrt{(a^2+b^2)^2}=\sqrt{2}(a^2+b^2)]



Now plug these values into the given formula above


*[Tex \LARGE \cos(\theta)=\frac{a^2+b^2}{\sqrt{2}(a^2+b^2)}]



Cancel out the common terms



*[Tex \LARGE \cos(\theta)=\frac{1}{\sqrt{2}}]




Now take the arccosine of both sides (to solve for theta)


*[Tex \LARGE \theta=\arccos{\left(\frac{1}{\sqrt{2}}\right)}]



Evaluate the arccosine of {{{1/sqrt(2)}}} to get {{{pi/4}}} radians (or 45 degrees). Note: use the unit circle


*[Tex \LARGE \theta=\frac{\pi}{4}] or *[Tex \LARGE \theta=45]




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Answer:


So the vector *[Tex \LARGE <a,b>] needs to rotate {{{pi/4}}} radians (or 45 degrees) in order to be parallel to the vector *[Tex \LARGE <a-b,a+b>]




Now if any of this is not making any sense, simply pick values for "a" and "b" and plug them in. So let's say that we choose {{{a=2}}} and {{{b=3}}}. So the first number would be {{{2+3i}}} and the second number would be {{{(2-3)+(2+3)i=-1+5i}}}


Now find the angle between the vectors <2,3> and <-1,5> and you'll find that it will be 45 degrees.