Express $z = \dfrac{3i}{\sqrt{2-i}} +1$ in the form $a + bi$, where $a, b \in\Bbb R$.

Question

Express $$z = \frac{3i}{\sqrt{2-i}} +1$$ in the form $a + bi$, where $a, b \in\Bbb R$.

I figure for this one I multiply by the conjugate of $\sqrt{2 +1}$? But I’m still struggle to achieve the form $a+bi$.

Answer 1

$\displaystyle z=\frac{3i}{\sqrt{2}-i+1}$

$\displaystyle =\frac{3i}{\sqrt{2}-i+1}\times\frac{\sqrt{2}+i+1}{\sqrt{2}+i+1}$

$\displaystyle =\frac{-3+(3+\sqrt{2})i}{(\sqrt{2}+1)^2+1}$

$\displaystyle =\frac{-3+(3+\sqrt{2})i}{4+2\sqrt{2}}$

Answer 2

Given $$\dfrac{3i}{\sqrt{2-i}}+1=\dfrac{3i+\sqrt{2-i}}{\sqrt{2-i}}$$ Now $$=\dfrac{3i+\sqrt{2-i}}{\sqrt{2-i}}\times \dfrac{\sqrt{2+i}}{\sqrt{2+i}}$$$$=\dfrac{3i\sqrt{2+i}+\sqrt{5}}{\sqrt{5}}$$ $$=\dfrac{3i\sqrt{2+i}}{\sqrt{5}}+\dfrac{\sqrt{5}}{\sqrt{5}}$$

Answer 3

You're gonna need to convert $\sqrt {2-i}$ to the form $\alpha + i \beta$.

To do that, convert to polar

$$ 2-i=r(\cos \theta + i \sin \theta)$$

So that

$$\sqrt{2-i}=\sqrt r \cdot (\cos \theta + i \sin \theta)^{1/2} = \sqrt r \cdot \left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right) $$

Some of the ingredients are:

$$r=\sqrt{2^2+(-1)^2}=\sqrt 5, \quad \cos \theta= \frac{2}{\sqrt 5}, \quad \sin \theta =\frac{-1}{\sqrt 5} $$

$\theta$ will be a negative angle, so that

$$\cos{\frac{\theta}{2}}=\sqrt{\frac{1+\cos \theta}{2}} \qquad \sin{\frac{\theta}{2}}=-\sqrt{\frac{1-\cos \theta}{2}}$$

Continuing

$$\cos{\frac{\theta}{2}}=\sqrt{\frac{1+\frac{2}{\sqrt 5}}{2}}= \sqrt{\frac{\sqrt 5+2}{2\sqrt 5}}=\frac{\sqrt{2+\sqrt 5}}{\sqrt 2 \sqrt r}$$

$$\sin{\frac{\theta}{2}}=-\sqrt{\frac{1-\frac{2}{\sqrt 5}}{2}}= -\sqrt{\frac{\sqrt 5-2}{2\sqrt 5}}=-\frac{\sqrt{-2+\sqrt 5}}{\sqrt 2 \sqrt r}$$

Making

$$\sqrt{2-i}=\left(\frac{\sqrt{2+\sqrt 5}}{\sqrt 2}\right)-i \left(\frac{\sqrt{-2+\sqrt 5}}{\sqrt 2 }\right) \in \alpha + i \beta$$

Answer 4

There is a question, to me, anyway, whether the expression is $\frac{3i}{\sqrt{2}- i}+ 1$ which is what daruma answered, or $\frac{3i}{\sqrt{2- i}}+ 1$. If it is the latter then $2- i= \sqrt{5}e^{-2i\pi/3}$ so that $\sqrt{2- i}= \sqrt[4]{5}e^{-i\pi/3}$. In "Cartesian form" that is $\sqrt[4]{5}(cos(\pi/3)- i sin(\pi/3))$. Its conjugate is $\sqrt[4]{5}(cos(\pi/3)+ i sin(\pi/3))$