Complex Numbers question - using the conjugate

Question

Compute the numerical value, give answer in the form $a+ib$ where $a$ and $b$ are real coefficients: $$\left(i+\frac{1}{1-2i}\right)^2$$

Answer 1

Hint:

$$i^2+2(i)\left(\dfrac 1{1-2i}\right)+\left(\dfrac 1{1-2i}\right)^2=-1+\dfrac{2i}{1-2i}+\dfrac {1}{(1-2i)^2}=$$$$-1+\dfrac{2i}{1-2i}\cdot \dfrac {1+2i}{1+2i}+\dfrac {1}{1-4i+4i^2}$$

$$-1+\dfrac{2i+4i^2}{1-4i^2}-\dfrac {1}{3+4i}=$$$$$$$$-1+\dfrac{-4+2i}{5}-\dfrac {1}{3+4i}\cdot \dfrac {3-4i}{3-4i}=-1-\dfrac 45+\dfrac {2i}5-\dfrac {3-4i}{9-16i^2}=$$$$$$$$-\dfrac{9}5+\dfrac{2i}5-\dfrac{3-4i}{25}=-\dfrac{45}{25}+\dfrac{10i}{25}-\dfrac {3}{25}+\dfrac{4i}{25}=-\dfrac{48}{25}+\dfrac {14}{25}i$$ Therefore, $a=-\dfrac{48}{25}$ and $b=\dfrac {14}{25}$

Answer 2

Hint:

Substitute $$\dfrac{1}{1-2i}=\dfrac{1}{1-2i}\dfrac{1+2i}{1+2i}=\dfrac{1+2i}{5}$$and proceed through

Answer 3

HINT

Use that

• $\frac{1}{1-2i}=\frac{1}{1-2i}\frac{1+2i}{1+2i}=\frac{1+2i}{5}$

and

• $(x+y)^2=x^2+2xy+y^2$