Find w=a + bi = square root of z , where a and b are real numbers.

i=sqrt(-1)

you can't find the square root of i because i is an imaginary number unless you put sqrt(i)

Find the square root of 60i-11 

$\begin{array}{rcll} z^2 &=& a+b\cdot i\\ z^2 &=& -11+60i \qquad a = -11 \qquad b = 60\\ z&=&\sqrt{-11+60i}\\\\ r &=& \sqrt{a^2+b^2} \\ r &=& \sqrt{(-11)^2+60^2}\\ r &=& 61\\\\ \tan{(\varphi)} &=& \frac{b}{a} \\ \tan{(\varphi)} &=& \frac{60}{-11} \qquad (\mathbf{II.})\\ \tan{(\varphi)} &=& -5.45454545455 \\ \varphi &=& \arctan{(-5.45454545455)} +180^\circ\\ \varphi &=& -79.6111421845 +180^\circ\\ \varphi &=& 100.388857815^\circ \\\\ z &=& \sqrt{r}\cdot \left[ \cos{( \frac{\varphi}{2} + k\cdot \frac{360^\circ}{2} )} + i \cdot \sin{ (\frac{\varphi}{2} + k\cdot \frac{360^\circ}{2}) } \right] \qquad k=0,1\\\\ z_0 &=& \sqrt{61}\cdot \left[ \cos{( \frac{100.388857815^\circ }{2} + 0\cdot \frac{360^\circ}{2} )} + i \cdot \sin{ (\frac{100.388857815^\circ }{2} + 0\cdot \frac{360^\circ}{2}) } \right]\\ z_0 &=& 7.81024967591\cdot \left[ \cos{( 50.1944289077^\circ )} + i \cdot \sin{ (50.1944289077^\circ) } \right]\\ z_0 &=& 7.81024967591\cdot \cos{( 50.1944289077^\circ )} + i \cdot 7.81024967591 \cdot \sin{ (50.1944289077^\circ) } \\ \mathbf{z_0} &\mathbf{=}& \mathbf{5 + 6\cdot i} \\\\ z_1 &=& \sqrt{61}\cdot \left[ \cos{( \frac{100.388857815^\circ }{2} + 1\cdot \frac{360^\circ}{2} )} + i \cdot \sin{ (\frac{100.388857815^\circ }{2} + 1\cdot \frac{360^\circ}{2}) } \right]\\ z_1 &=& 7.81024967591\cdot \left[ \cos{( 230.194428908^\circ )} + i \cdot \sin{ (230.194428908^\circ) } \right]\\ z_1 &=& 7.81024967591\cdot \cos{( 230.194428908^\circ )} + i \cdot 7.81024967591\cdot \sin{ (230.194428908^\circ) } \\ \mathbf{z_1 }&\mathbf{=}&\mathbf{ -5 -6\cdot i }\\ \end{array}$