Simplfy: `(5+3i)/(1+2i)+(3+7i)/(1-2i)`

Express the result in the form x+iy, where x,y are real number i=sqrt(-1) : (i) (5+9i)-:(-3+4i) (ii) [(sqrt(5)+(i)/(2))(sqrt(5)-2i)]-:(6+5i) (iii) ((1-i)(2-i)(3-i))/(1+i) (iv) (1+3i)/((1-2i)^(2))

Simplify : (i)" "(-2i)((1)/(6)i)" "(ii)" "(-i)(3i)((-1)/(6)i)^(3)" "(iii)" "4 sqrt(-4)+5 sqrt(-9)-3 sqrt(-16)

Prove that (3+i)/(1+2i)+(3-i)/(1-2i) is a real number.

((1+2i)^(3))/((1+i)(2-i))

Convert the following in the polar form : ( i ) (1+7i)/((2-i)^(2)) (ii) (1+3i)/(1-2i)

Express the following complex number in the polar form: (i) (1+7i)/((2-i)^(2))( ii) (1+3i)/(1-2i)

Simplify : {:((i),3(6+6i)+i(6+6i),(ii),(1-i)-(-3+6i)),((iii),((1)/(3)-(2)/(3)i)-(4+(3)/(2)i),(iv),{((1)/(5)+(7)/(5)i)-(6+(1)/(5)i)}-((-4)/(5)+i)):}

Express each of the following in the form (a + ib) and find its multiplicative inverse : (i) (1+2i)/(1-3i) (ii) ((1+7i))/((2-i)^(2)) (iii) (-4)/((1+i sqrt(3)))

" If "z=((1+i)(1+2i)(1+3i))/((1-i)(2-i)(3-i))" then the principal argument of "z"

Express the result in the form x+iy, where x,y are real number i=sqrt(-1) : (i) (2-3i)/(4-i) (ii) (2+3i)/(-5-4i) (iii) (1+i)/(3+i) (iv) (3+2i)/(4-3i)