Converting complex numbers to their $x+iy$ form

Question

I was wondering if someone could explain these conversions to me and how they turn into each other

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$$\frac{1}{1+i}=0.5+0.5i$$

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$$i^4=(1-0i)$$

I thought the above one would be $0+1i$?

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$$(i^2)+2i+1= 0+2i$$

This one I'm quite confused about.

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Thanks

Answer 1

Hint

Clever multiplication by $1$ and manipulation is all it asks for.

$$\dfrac{1}{1+i}=\dfrac{1}{1+i}\cdot\dfrac{1-i}{1-i}$$

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Note that $i^2$ is defined as $-1$.

$$i^4=(i^2)^2=(-1)^2$$

$$i^2+2i+1=-1+1+2i=2i$$

Answer 2

For the first one, multiply the numerator and denominator by the conjugate of the latter. (The conjugate $\bar{z}$ of a complex number $z = x+iy$ is just the same as $z$ with the sign flipped on the imaginary term, i.e. $\bar{z} = x-iy.$)

For the second one, recall $i^2 = -1$ in the complex numbers, and thus $(i^2)^2 = (-1)^2 = 1$.

For the third one, again note $i^2 = -1$ and the result appears quite nicely.