Answer:
[tex]\sqrt{1 + \sqrt{3} i}=\pm (1.58+i 0.548)[/tex]
Step-by-step explanation:
Given
z = 1 + √3 i
Let [tex]\sqrt{1+\sqrt(3) i}=p+iq[/tex]
Squaring both sides
[tex]1+\sqrt(3) i=p^2-q^2+2ipq[/tex]
Comparing real and imaginary part
Re(LHS)=Re(RHS)
[tex]1=p^2-q^2[/tex]...........................(1)
comparing Im(LHS)=Im(RHS)
√3=2pq
[tex]q=\frac{\sqrt{3}}{2p}[/tex]
Substitute q in equation (1)
[tex]1=p^2-(\frac{\sqrt{3}}{2p})^2[/tex]
[tex]p^4-p^2-0.75=0[/tex]
Let [tex]x=p^2[/tex]
[tex]x^2-x-0.75=0[/tex]
[tex]x=\dfrac{1\pm \sqrt{1^2+4\times 0.75}}{2}[/tex]
[tex]x=\frac{1\pm 4}{2}[/tex]
we take only Positive value because [tex]p^2=x[/tex]
x=2.5
[tex]p^2=2.5[/tex]
thus [tex]p=\pm 1.58[/tex]
[tex]q=\pm 0.548[/tex]
thus,
[tex]\sqrt{1 + \sqrt{3} i}=\pm (1.58+i 0.548)[/tex]