super

Note that if you call super on a child class it is redirected to the prototype as shown below:

class Base {
    log() { console.log('hello world'); }
}

class Child extends Base {
    log() { super.log() };
}

generates:

var Base = (function () {
    function Base() {
    }
    Base.prototype.log = function () { console.log('hello world'); };
    return Base;
})();
var Child = (function (_super) {
    __extends(Child, _super);
    function Child() {
        _super.apply(this, arguments);
    }
    Child.prototype.log = function () { _super.prototype.log.call(this); };
    return Child;
})(Base);

Notice _super.prototype.log.call(this).

This means that you cannot use super on member properties. Instead you should just use this.

class Base {
    log = () => { console.log('hello world'); }
}

class Child extends Base {
    logWorld() { this.log() };
}

Notice since there is only one this shared between the Base and the Child class you need to use different names (here log and logWorld).

Also Note that TypeScript will warn you if you try to misuse super:

module quz {
    class Base {
        log = () => { console.log('hello world'); }
    }

    class Child extends Base {
        // ERROR : only `public` and `protected` methods of base class are accessible via `super`
        logWorld() { super.log() };
    }
}